Chapter 8 Binomial Theorem

Given expression:

$\left( x^{2} +\frac{3}{x}\right)^{4}$

This is in the form $(a+b)^n$, where $a = x^2$, $b = \frac{3}{x}$, and $n = 4$.

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We can expand this expression using the Binomial Theorem, which states that:

$(a+b)^n = \binom{n}{0}a^n b^0 + \binom{n}{1}a^{n-1} b^1 + \binom{n}{2}a^{n-2} b^2 + ... + \binom{n}{n-1}a^1 b^{n-1} + \binom{n}{n}a^0 b^n$

For $n=4$, the expansion is:

$(a+b)^4 = \binom{4}{0}a^4 b^0 + \binom{4}{1}a^3 b^1 + \binom{4}{2}a^2 b^2 + \binom{4}{3}a^1 b^3 + \binom{4}{4}a^0 b^4$

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Substituting $a = x^2$ and $b = \frac{3}{x}$ into the expansion:

$\left( x^{2} +\frac{3}{x}\right)^{4} = \binom{4}{0}(x^2)^4 \left(\frac{3}{x}\right)^0 + \binom{4}{1}(x^2)^3 \left(\frac{3}{x}\right)^1 + \binom{4}{2}(x^2)^2 \left(\frac{3}{x}\right)^2 + \binom{4}{3}(x^2)^1 \left(\frac{3}{x}\right)^3 + \binom{4}{4}(x^2)^0 \left(\frac{3}{x}\right)^4$

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Now, we calculate each term:

Term 1: $\binom{4}{0}(x^2)^4 \left(\frac{3}{x}\right)^0 = 1 \times x^{2 \times 4} \times 1 = 1 \times x^8 \times 1 = x^8$

Term 2: $\binom{4}{1}(x^2)^3 \left(\frac{3}{x}\right)^1 = 4 \times x^{2 \times 3} \times \frac{3}{x} = 4 \times x^6 \times \frac{3}{x} = 12 \frac{x^6}{x} = 12x^{6-1} = 12x^5$

Term 3: $\binom{4}{2}(x^2)^2 \left(\frac{3}{x}\right)^2 = 6 \times x^{2 \times 2} \times \frac{3^2}{x^2} = 6 \times x^4 \times \frac{9}{x^2} = 54 \frac{x^4}{x^2} = 54x^{4-2} = 54x^2$

Term 4: $\binom{4}{3}(x^2)^1 \left(\frac{3}{x}\right)^3 = 4 \times x^2 \times \frac{3^3}{x^3} = 4 \times x^2 \times \frac{27}{x^3} = 108 \frac{x^2}{x^3} = 108x^{2-3} = 108x^{-1} = \frac{108}{x}$

Term 5: $\binom{4}{4}(x^2)^0 \left(\frac{3}{x}\right)^4 = 1 \times 1 \times \frac{3^4}{x^4} = 1 \times 1 \times \frac{81}{x^4} = \frac{81}{x^4}$

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Adding the terms together, we get the expanded form:

$\left( x^{2} +\frac{3}{x}\right)^{4} = x^8 + 12x^5 + 54x^2 + \frac{108}{x} + \frac{81}{x^4}$

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The expanded form is:

$\mathbf{\left( x^{2} +\frac{3}{x}\right)^{4} = x^8 + 12x^5 + 54x^2 + \frac{108}{x} + \frac{81}{x^4}}$

Given expression:

$(98)^5$

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We can write 98 as $100 - 2$. So, we need to compute $(100 - 2)^5$.

This is in the form $(a-b)^n$, where $a = 100$, $b = 2$, and $n = 5$.

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We can expand this expression using the Binomial Theorem:

$(a-b)^n = \binom{n}{0}a^n b^0 - \binom{n}{1}a^{n-1} b^1 + \binom{n}{2}a^{n-2} b^2 - \binom{n}{3}a^{n-3} b^3 + ... + (-1)^n \binom{n}{n}a^0 b^n$

For $n=5$, the expansion is:

$(100-2)^5 = \binom{5}{0}(100)^5 (2)^0 - \binom{5}{1}(100)^4 (2)^1 + \binom{5}{2}(100)^3 (2)^2 - \binom{5}{3}(100)^2 (2)^3 + \binom{5}{4}(100)^1 (2)^4 - \binom{5}{5}(100)^0 (2)^5$

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Now, we calculate each term:

Term 1: $\binom{5}{0}(100)^5 (2)^0 = 1 \times 10^{10} \times 1 = 10,000,000,000$

Term 2: $-\binom{5}{1}(100)^4 (2)^1 = -5 \times 10^8 \times 2 = -10 \times 100,000,000 = -1,000,000,000$

Term 3: $+\binom{5}{2}(100)^3 (2)^2 = +10 \times 10^6 \times 4 = +40 \times 1,000,000 = +40,000,000$

Term 4: $-\binom{5}{3}(100)^2 (2)^3 = -10 \times 10^4 \times 8 = -80 \times 10,000 = -800,000$

Term 5: $+\binom{5}{4}(100)^1 (2)^4 = +5 \times 100 \times 16 = +500 \times 16 = +8,000$

Term 6: $-\binom{5}{5}(100)^0 (2)^5 = -1 \times 1 \times 32 = -32$

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Adding the terms together:

$(98)^5 = 10,000,000,000 - 1,000,000,000 + 40,000,000 - 800,000 + 8,000 - 32$

$(98)^5 = 9,000,000,000 + 40,000,000 - 800,000 + 8,000 - 32$

$(98)^5 = 9,040,000,000 - 800,000 + 8,000 - 32$

$(98)^5 = 9,039,200,000 + 8,000 - 32$

$(98)^5 = 9,039,208,000 - 32$

$(98)^5 = 9,039,207,968$

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The computed value is:

$(98)^5 = \mathbf{9,039,207,968}$

We need to compare the values of $(1.01)^{1000000}$ and $10,000$.

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Let the first number be $A = (1.01)^{1000000}$.

We can write $1.01$ as $1 + 0.01$.

So, $A = (1 + 0.01)^{1000000}$.

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We can expand this expression using the Binomial Theorem for $(1+x)^n$, which is given by:

$(1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \binom{n}{3}x^3 + ... + \binom{n}{n}x^n$

In this case, $x = 0.01$ and $n = 1000000$.

Expanding $(1 + 0.01)^{1000000}$:

$(1 + 0.01)^{1000000} = \binom{1000000}{0} + \binom{1000000}{1}(0.01) + \binom{1000000}{2}(0.01)^2 + \binom{1000000}{3}(0.01)^3 + ...$

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Let's calculate the first two terms of the expansion:

The first term is $\binom{1000000}{0} = 1$.

The second term is $\binom{1000000}{1}(0.01) = 1000000 \times 0.01 = 10000$.

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So, the expansion starts with $1 + 10000 + \text{ subsequent terms}$.

The subsequent terms are $\binom{1000000}{2}(0.01)^2$, $\binom{1000000}{3}(0.01)^3$, and so on.

Since $n = 1000000$ is a positive integer and $x = 0.01$ is a positive number, all the terms in the binomial expansion $\binom{n}{k}x^k$ for $k \geq 2$ will be positive.

Therefore, the sum of the subsequent terms is a positive value.

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This means:

$(1.01)^{1000000} = 1 + 10000 + (\text{a positive value})$

$(1.01)^{1000000} = 10001 + (\text{a positive value})$

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From this, it is clear that $(1.01)^{1000000}$ is greater than $10001$.

Since $10001 > 10000$, we can conclude that $(1.01)^{1000000}$ is larger than $10000$.

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Thus, $\mathbf{(1.01)^{1000000} \text{ is larger than } 10,000}$.

To Prove:

The expression $6^n - 5n$ leaves a remainder of 1 when divided by 25, for any positive integer $n$.

This is equivalent to proving that $6^n - 5n \equiv 1 \pmod{25}$, or $6^n - 5n - 1$ is divisible by 25.

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Proof:

Consider the expression $6^n$. We can write $6$ as $1 + 5$.

So, $6^n = (1 + 5)^n$.

Using the Binomial Theorem for a positive integer $n$, the expansion of $(1+x)^n$ is given by:

$(1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \binom{n}{3}x^3 + ... + \binom{n}{n}x^n$

Substitute $x = 5$ into the binomial expansion of $(1+5)^n$:

$6^n = (1+5)^n = \binom{n}{0}1^n 5^0 + \binom{n}{1}1^{n-1} 5^1 + \binom{n}{2}1^{n-2} 5^2 + \binom{n}{3}1^{n-3} 5^3 + ... + \binom{n}{n}1^0 5^n$

$6^n = \binom{n}{0} + \binom{n}{1}5 + \binom{n}{2}5^2 + \binom{n}{3}5^3 + ... + \binom{n}{n}5^n$

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We know that $\binom{n}{0} = 1$ and $\binom{n}{1} = n$. Also, $5^2 = 25$, $5^3 = 125$, and so on.

So, the expansion becomes:

$6^n = 1 + n \times 5 + \binom{n}{2} \times 25 + \binom{n}{3} \times 125 + ... + \binom{n}{n} \times 5^n$

$6^n = 1 + 5n + 25\binom{n}{2} + 125\binom{n}{3} + ... + 5^n\binom{n}{n}$

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Now, consider the expression $6^n - 5n$. Substitute the expansion of $6^n$ into this expression:

$6^n - 5n = (1 + 5n + 25\binom{n}{2} + 125\binom{n}{3} + ... + 5^n\binom{n}{n}) - 5n$

$6^n - 5n = 1 + (5n - 5n) + 25\binom{n}{2} + 125\binom{n}{3} + ... + 5^n\binom{n}{n}$

$6^n - 5n = 1 + 25\binom{n}{2} + 125\binom{n}{3} + ... + 5^n\binom{n}{n}$

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For $n=1$, the expansion stops at $\binom{1}{1}5^1$. The terms $\binom{n}{k}$ for $k \geq 2$ are zero.

If $n=1$: $6^1 - 5(1) = 6 - 5 = 1$. When 1 is divided by 25, the remainder is 1.

Using the formula: $6^1 - 5(1) = 1 + 25\binom{1}{2} + 125\binom{1}{3} + ... = 1 + 25(0) + 125(0) + ... = 1$. This confirms the formula holds for $n=1$.

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For $n \geq 2$, every term from $25\binom{n}{2}$ onwards has a factor of 25 (since $5^k$ for $k \geq 2$ is divisible by 25).

We can factor out 25 from all terms except the first one:

$6^n - 5n = 1 + 25\left[\binom{n}{2} + \frac{125}{25}\binom{n}{3} + ... + \frac{5^n}{25}\binom{n}{n}\right]$

$6^n - 5n = 1 + 25\left[\binom{n}{2} + 5\binom{n}{3} + ... + 5^{n-2}\binom{n}{n}\right]$

Let $K = \binom{n}{2} + 5\binom{n}{3} + ... + 5^{n-2}\binom{n}{n}$. Since $n$ is a positive integer, $\binom{n}{k}$ are integers, and powers of 5 are integers, $K$ is an integer for $n \geq 2$. For $n=1$, $K$ is an empty sum, which is 0. For $n \geq 1$, $K$ is an integer.

So, for any positive integer $n$, we can write:

$6^n - 5n = 1 + 25K$

where $K$ is an integer.

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Rearranging the equation, we get:

$6^n - 5n - 1 = 25K$

Since $K$ is an integer, $25K$ is a multiple of 25. This means that $6^n - 5n - 1$ is divisible by 25.

By the definition of divisibility, this implies that when $6^n - 5n$ is divided by 25, the remainder is 1.

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Conclusion:

Using the binomial theorem, we have shown that $6^n - 5n$ can be expressed in the form $1 + 25K$ for some integer $K$ (for any positive integer $n$). This proves that $6^n - 5n$ always leaves a remainder of 1 when divided by 25.

Hence Proved.

Given expression:

$(1 - 2x)^5$

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This expression is in the form $(a+b)^n$, where $a = 1$, $b = -2x$, and $n = 5$.

We use the Binomial Theorem for expansion:

$(a+b)^n = \binom{n}{0}a^n b^0 + \binom{n}{1}a^{n-1} b^1 + \binom{n}{2}a^{n-2} b^2 + \binom{n}{3}a^{n-3} b^3 + ... + \binom{n}{n}a^0 b^n$

For $n=5$, the expansion of $(1 - 2x)^5$ is:

$(1 - 2x)^5 = \binom{5}{0}(1)^5 (-2x)^0 + \binom{5}{1}(1)^4 (-2x)^1 + \binom{5}{2}(1)^3 (-2x)^2 + \binom{5}{3}(1)^2 (-2x)^3 + \binom{5}{4}(1)^1 (-2x)^4 + \binom{5}{5}(1)^0 (-2x)^5$

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Now, we calculate each term:

Term 1: $\binom{5}{0}(1)^5 (-2x)^0 = 1 \times 1 \times 1 = 1$

Term 2: $\binom{5}{1}(1)^4 (-2x)^1 = 5 \times 1 \times (-2x) = -10x$

Term 3: $\binom{5}{2}(1)^3 (-2x)^2 = 10 \times 1 \times (4x^2) = 40x^2$

Term 4: $\binom{5}{3}(1)^2 (-2x)^3 = 10 \times 1 \times (-8x^3) = -80x^3$

Term 5: $\binom{5}{4}(1)^1 (-2x)^4 = 5 \times 1 \times (16x^4) = 80x^4$

Term 6: $\binom{5}{5}(1)^0 (-2x)^5 = 1 \times 1 \times (-32x^5) = -32x^5$

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Adding the terms together, we get the expanded form:

$(1 - 2x)^5 = 1 + (-10x) + 40x^2 + (-80x^3) + 80x^4 + (-32x^5)$

$(1 - 2x)^5 = 1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5$

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The expanded form is:

$\mathbf{(1 - 2x)^5 = 1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5}$

Given expression:

$\left( \frac{2}{x}-\frac{x}{2} \right)^{5}$

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This expression is in the form $(a+b)^n$, where $a = \frac{2}{x}$, $b = -\frac{x}{2}$, and $n = 5$.

We use the Binomial Theorem for expansion:

$(a+b)^n = \binom{n}{0}a^n b^0 + \binom{n}{1}a^{n-1} b^1 + \binom{n}{2}a^{n-2} b^2 + \binom{n}{3}a^{n-3} b^3 + ... + \binom{n}{n}a^0 b^n$

For $n=5$, the expansion of $\left( \frac{2}{x}-\frac{x}{2} \right)^{5}$ is:

$\binom{5}{0}\left(\frac{2}{x}\right)^5 \left(-\frac{x}{2}\right)^0 + \binom{5}{1}\left(\frac{2}{x}\right)^4 \left(-\frac{x}{2}\right)^1 + \binom{5}{2}\left(\frac{2}{x}\right)^3 \left(-\frac{x}{2}\right)^2 + \binom{5}{3}\left(\frac{2}{x}\right)^2 \left(-\frac{x}{2}\right)^3 + \binom{5}{4}\left(\frac{2}{x}\right)^1 \left(-\frac{x}{2}\right)^4 + \binom{5}{5}\left(\frac{2}{x}\right)^0 \left(-\frac{x}{2}\right)^5$

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Now, we calculate each term:

Term 1: $\binom{5}{0}\left(\frac{2}{x}\right)^5 \left(-\frac{x}{2}\right)^0 = 1 \times \frac{2^5}{x^5} \times 1 = \frac{32}{x^5}$

Term 2: $\binom{5}{1}\left(\frac{2}{x}\right)^4 \left(-\frac{x}{2}\right)^1 = 5 \times \frac{2^4}{x^4} \times \left(-\frac{x}{2}\right) = 5 \times \frac{16}{x^4} \times \left(-\frac{x}{2}\right) = -5 \times \frac{16x}{2x^4} = -5 \times \frac{8}{x^3} = -\frac{40}{x^3}$

Term 3: $\binom{5}{2}\left(\frac{2}{x}\right)^3 \left(-\frac{x}{2}\right)^2 = 10 \times \frac{2^3}{x^3} \times \frac{x^2}{2^2} = 10 \times \frac{8}{x^3} \times \frac{x^2}{4} = 10 \times \frac{8x^2}{4x^3} = 10 \times \frac{2}{x} = \frac{20}{x}$

Term 4: $\binom{5}{3}\left(\frac{2}{x}\right)^2 \left(-\frac{x}{2}\right)^3 = 10 \times \frac{2^2}{x^2} \times \left(-\frac{x^3}{2^3}\right) = 10 \times \frac{4}{x^2} \times \left(-\frac{x^3}{8}\right) = -10 \times \frac{4x^3}{8x^2} = -10 \times \frac{x}{2} = -5x$

Term 5: $\binom{5}{4}\left(\frac{2}{x}\right)^1 \left(-\frac{x}{2}\right)^4 = 5 \times \frac{2}{x} \times \frac{x^4}{2^4} = 5 \times \frac{2}{x} \times \frac{x^4}{16} = 5 \times \frac{2x^4}{16x} = 5 \times \frac{x^3}{8} = \frac{5x^3}{8}$

Term 6: $\binom{5}{5}\left(\frac{2}{x}\right)^0 \left(-\frac{x}{2}\right)^5 = 1 \times 1 \times \left(-\frac{x^5}{2^5}\right) = -\frac{x^5}{32}$

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Adding the terms together, we get the expanded form:

$\left( \frac{2}{x}-\frac{x}{2} \right)^{5} = \frac{32}{x^5} - \frac{40}{x^3} + \frac{20}{x} - 5x + \frac{5x^3}{8} - \frac{x^5}{32}$

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The expanded form is:

$\mathbf{\left( \frac{2}{x}-\frac{x}{2} \right)^{5} = \frac{32}{x^5} - \frac{40}{x^3} + \frac{20}{x} - 5x + \frac{5x^3}{8} - \frac{x^5}{32}}$

Given expression:

$(2x - 3)^6$

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This expression is in the form $(a+b)^n$, where $a = 2x$, $b = -3$, and $n = 6$.

We use the Binomial Theorem for expansion:

$(a+b)^n = \binom{n}{0}a^n b^0 + \binom{n}{1}a^{n-1} b^1 + \binom{n}{2}a^{n-2} b^2 + \binom{n}{3}a^{n-3} b^3 + ... + \binom{n}{n}a^0 b^n$

For $n=6$, the expansion of $(2x - 3)^6$ is:

$\binom{6}{0}(2x)^6 (-3)^0 + \binom{6}{1}(2x)^5 (-3)^1 + \binom{6}{2}(2x)^4 (-3)^2 + \binom{6}{3}(2x)^3 (-3)^3 + \binom{6}{4}(2x)^2 (-3)^4 + \binom{6}{5}(2x)^1 (-3)^5 + \binom{6}{6}(2x)^0 (-3)^6$

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Now, we calculate each term:

Term 1: $\binom{6}{0}(2x)^6 (-3)^0 = 1 \times (64x^6) \times 1 = 64x^6$

Term 2: $\binom{6}{1}(2x)^5 (-3)^1 = 6 \times (32x^5) \times (-3) = 6 \times (-96x^5) = -576x^5$

Term 3: $\binom{6}{2}(2x)^4 (-3)^2 = 15 \times (16x^4) \times 9 = 15 \times (144x^4) = 2160x^4$

Term 4: $\binom{6}{3}(2x)^3 (-3)^3 = 20 \times (8x^3) \times (-27) = 20 \times (-216x^3) = -4320x^3$

Term 5: $\binom{6}{4}(2x)^2 (-3)^4 = 15 \times (4x^2) \times 81 = 15 \times (324x^2) = 4860x^2$

Term 6: $\binom{6}{5}(2x)^1 (-3)^5 = 6 \times (2x) \times (-243) = 6 \times (-486x) = -2916x$

Term 7: $\binom{6}{6}(2x)^0 (-3)^6 = 1 \times 1 \times 729 = 729$

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Adding the terms together, we get the expanded form:

$(2x - 3)^6 = 64x^6 - 576x^5 + 2160x^4 - 4320x^3 + 4860x^2 - 2916x + 729$

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The expanded form is:

$\mathbf{(2x - 3)^6 = 64x^6 - 576x^5 + 2160x^4 - 4320x^3 + 4860x^2 - 2916x + 729}$

Given expression:

$\left( \frac{x}{3}+\frac{1}{x} \right)^{5}$

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This expression is in the form $(a+b)^n$, where $a = \frac{x}{3}$, $b = \frac{1}{x}$, and $n = 5$.

We use the Binomial Theorem for expansion:

$(a+b)^n = \binom{n}{0}a^n b^0 + \binom{n}{1}a^{n-1} b^1 + \binom{n}{2}a^{n-2} b^2 + \binom{n}{3}a^{n-3} b^3 + ... + \binom{n}{n}a^0 b^n$

For $n=5$, the expansion of $\left( \frac{x}{3}+\frac{1}{x} \right)^{5}$ is:

$\binom{5}{0}\left(\frac{x}{3}\right)^5 \left(\frac{1}{x}\right)^0 + \binom{5}{1}\left(\frac{x}{3}\right)^4 \left(\frac{1}{x}\right)^1 + \binom{5}{2}\left(\frac{x}{3}\right)^3 \left(\frac{1}{x}\right)^2 + \binom{5}{3}\left(\frac{x}{3}\right)^2 \left(\frac{1}{x}\right)^3 + \binom{5}{4}\left(\frac{x}{3}\right)^1 \left(\frac{1}{x}\right)^4 + \binom{5}{5}\left(\frac{x}{3}\right)^0 \left(\frac{1}{x}\right)^5$

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Now, we calculate each term:

Term 1: $\binom{5}{0}\left(\frac{x}{3}\right)^5 \left(\frac{1}{x}\right)^0 = 1 \times \frac{x^5}{3^5} \times 1 = \frac{x^5}{243}$

Term 2: $\binom{5}{1}\left(\frac{x}{3}\right)^4 \left(\frac{1}{x}\right)^1 = 5 \times \frac{x^4}{3^4} \times \frac{1}{x} = 5 \times \frac{x^4}{81} \times \frac{1}{x} = 5 \times \frac{x^{4-1}}{81} = 5 \times \frac{x^3}{81} = \frac{5x^3}{81}$

Term 3: $\binom{5}{2}\left(\frac{x}{3}\right)^3 \left(\frac{1}{x}\right)^2 = 10 \times \frac{x^3}{3^3} \times \frac{1}{x^2} = 10 \times \frac{x^3}{27} \times \frac{1}{x^2} = 10 \times \frac{x^{3-2}}{27} = 10 \times \frac{x}{27} = \frac{10x}{27}$

Term 4: $\binom{5}{3}\left(\frac{x}{3}\right)^2 \left(\frac{1}{x}\right)^3 = 10 \times \frac{x^2}{3^2} \times \frac{1}{x^3} = 10 \times \frac{x^2}{9} \times \frac{1}{x^3} = 10 \times \frac{x^{2-3}}{9} = 10 \times \frac{x^{-1}}{9} = 10 \times \frac{1}{9x} = \frac{10}{9x}$

Term 5: $\binom{5}{4}\left(\frac{x}{3}\right)^1 \left(\frac{1}{x}\right)^4 = 5 \times \frac{x}{3} \times \frac{1}{x^4} = 5 \times \frac{x^{1-4}}{3} = 5 \times \frac{x^{-3}}{3} = 5 \times \frac{1}{3x^3} = \frac{5}{3x^3}$

Term 6: $\binom{5}{5}\left(\frac{x}{3}\right)^0 \left(\frac{1}{x}\right)^5 = 1 \times 1 \times \frac{1}{x^5} = \frac{1}{x^5}$

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Adding the terms together, we get the expanded form:

$\left( \frac{x}{3}+\frac{1}{x} \right)^{5} = \frac{x^5}{243} + \frac{5x^3}{81} + \frac{10x}{27} + \frac{10}{9x} + \frac{5}{3x^3} + \frac{1}{x^5}$

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The expanded form is:

$\mathbf{\left( \frac{x}{3}+\frac{1}{x} \right)^{5} = \frac{x^5}{243} + \frac{5x^3}{81} + \frac{10x}{27} + \frac{10}{9x} + \frac{5}{3x^3} + \frac{1}{x^5}}$

Given expression:

$\left( x+\frac{1}{x} \right)^{6}$

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This expression is in the form $(a+b)^n$, where $a = x$, $b = \frac{1}{x}$, and $n = 6$.

We use the Binomial Theorem for expansion:

$(a+b)^n = \binom{n}{0}a^n b^0 + \binom{n}{1}a^{n-1} b^1 + \binom{n}{2}a^{n-2} b^2 + ... + \binom{n}{n}a^0 b^n$

For $n=6$, the expansion of $\left( x+\frac{1}{x} \right)^{6}$ is:

$\binom{6}{0}(x)^6 \left(\frac{1}{x}\right)^0 + \binom{6}{1}(x)^5 \left(\frac{1}{x}\right)^1 + \binom{6}{2}(x)^4 \left(\frac{1}{x}\right)^2 + \binom{6}{3}(x)^3 \left(\frac{1}{x}\right)^3 + \binom{6}{4}(x)^2 \left(\frac{1}{x}\right)^4 + \binom{6}{5}(x)^1 \left(\frac{1}{x}\right)^5 + \binom{6}{6}(x)^0 \left(\frac{1}{x}\right)^6$

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Now, we calculate each term:

Term 1: $\binom{6}{0}(x)^6 \left(\frac{1}{x}\right)^0 = 1 \times x^6 \times 1 = x^6$

Term 2: $\binom{6}{1}(x)^5 \left(\frac{1}{x}\right)^1 = 6 \times x^5 \times \frac{1}{x} = 6 \times x^{5-1} = 6x^4$

Term 3: $\binom{6}{2}(x)^4 \left(\frac{1}{x}\right)^2 = 15 \times x^4 \times \frac{1}{x^2} = 15 \times x^{4-2} = 15x^2$

Term 4: $\binom{6}{3}(x)^3 \left(\frac{1}{x}\right)^3 = 20 \times x^3 \times \frac{1}{x^3} = 20 \times x^{3-3} = 20 \times x^0 = 20 \times 1 = 20$

Term 5: $\binom{6}{4}(x)^2 \left(\frac{1}{x}\right)^4 = 15 \times x^2 \times \frac{1}{x^4} = 15 \times x^{2-4} = 15x^{-2} = \frac{15}{x^2}$

Term 6: $\binom{6}{5}(x)^1 \left(\frac{1}{x}\right)^5 = 6 \times x^1 \times \frac{1}{x^5} = 6 \times x^{1-5} = 6x^{-4} = \frac{6}{x^4}$

Term 7: $\binom{6}{6}(x)^0 \left(\frac{1}{x}\right)^6 = 1 \times 1 \times \frac{1}{x^6} = \frac{1}{x^6}$

---

Adding the terms together, we get the expanded form:

$\left( x+\frac{1}{x} \right)^{6} = x^6 + 6x^4 + 15x^2 + 20 + \frac{15}{x^2} + \frac{6}{x^4} + \frac{1}{x^6}$

---

The expanded form is:

$\mathbf{\left( x+\frac{1}{x} \right)^{6} = x^6 + 6x^4 + 15x^2 + 20 + \frac{15}{x^2} + \frac{6}{x^4} + \frac{1}{x^6}}$

Given expression:

$(96)^3$

---

We can write 96 as $100 - 4$. So, we need to evaluate $(100 - 4)^3$.

This is in the form $(a-b)^n$, where $a = 100$, $b = 4$, and $n = 3$.

---

We use the Binomial Theorem for expansion:

$(a-b)^n = \binom{n}{0}a^n b^0 - \binom{n}{1}a^{n-1} b^1 + \binom{n}{2}a^{n-2} b^2 - ... + (-1)^n \binom{n}{n}a^0 b^n$

For $n=3$, the expansion of $(100 - 4)^3$ is:

$(100 - 4)^3 = \binom{3}{0}(100)^3 (4)^0 - \binom{3}{1}(100)^2 (4)^1 + \binom{3}{2}(100)^1 (4)^2 - \binom{3}{3}(100)^0 (4)^3$

---

Now, we calculate each term:

Term 1: $\binom{3}{0}(100)^3 (4)^0 = 1 \times (1000000) \times 1 = 1000000$

Term 2: $-\binom{3}{1}(100)^2 (4)^1 = -3 \times (10000) \times 4 = -3 \times 40000 = -120000$

Term 3: $+\binom{3}{2}(100)^1 (4)^2 = +3 \times 100 \times 16 = +300 \times 16 = +4800$

Term 4: $-\binom{3}{3}(100)^0 (4)^3 = -1 \times 1 \times 64 = -64$

---

Adding the terms together:

$(96)^3 = 1000000 - 120000 + 4800 - 64$

$(96)^3 = 880000 + 4800 - 64$

$(96)^3 = 884800 - 64$

$(96)^3 = 884736$

---

The evaluated value is:

$(96)^3 = \mathbf{884736}$

Given expression:

$(102)^5$

---

We can write 102 as $100 + 2$. So, we need to evaluate $(100 + 2)^5$.

This is in the form $(a+b)^n$, where $a = 100$, $b = 2$, and $n = 5$.

---

We use the Binomial Theorem for expansion:

$(a+b)^n = \binom{n}{0}a^n b^0 + \binom{n}{1}a^{n-1} b^1 + \binom{n}{2}a^{n-2} b^2 + \binom{n}{3}a^{n-3} b^3 + ... + \binom{n}{n}a^0 b^n$

For $n=5$, the expansion of $(100 + 2)^5$ is:

$(100 + 2)^5 = \binom{5}{0}(100)^5 (2)^0 + \binom{5}{1}(100)^4 (2)^1 + \binom{5}{2}(100)^3 (2)^2 + \binom{5}{3}(100)^2 (2)^3 + \binom{5}{4}(100)^1 (2)^4 + \binom{5}{5}(100)^0 (2)^5$

---

Now, we calculate each term:

Term 1: $\binom{5}{0}(100)^5 (2)^0 = 1 \times 10^{10} \times 1 = 10,000,000,000$

Term 2: $\binom{5}{1}(100)^4 (2)^1 = 5 \times 10^8 \times 2 = 10 \times 100,000,000 = 1,000,000,000$

Term 3: $\binom{5}{2}(100)^3 (2)^2 = 10 \times 10^6 \times 4 = 40 \times 1,000,000 = 40,000,000$

Term 4: $\binom{5}{3}(100)^2 (2)^3 = 10 \times 10^4 \times 8 = 80 \times 10,000 = 800,000$

Term 5: $\binom{5}{4}(100)^1 (2)^4 = 5 \times 100 \times 16 = 500 \times 16 = 8,000$

Term 6: $\binom{5}{5}(100)^0 (2)^5 = 1 \times 1 \times 32 = 32$

---

Adding the terms together:

$(102)^5 = 10,000,000,000 + 1,000,000,000 + 40,000,000 + 800,000 + 8,000 + 32$

$(102)^5 = 11,000,000,000 + 40,000,000 + 800,000 + 8,000 + 32$

$(102)^5 = 11,040,000,000 + 800,000 + 8,000 + 32$

$(102)^5 = 11,040,800,000 + 8,000 + 32$

$(102)^5 = 11,040,808,000 + 32$

$(102)^5 = 11,040,808,032$

---

The evaluated value is:

$(102)^5 = \mathbf{11,040,808,032}$

Given expression:

$(101)^4$

---

We can write 101 as $100 + 1$. So, we need to evaluate $(100 + 1)^4$.

This is in the form $(a+b)^n$, where $a = 100$, $b = 1$, and $n = 4$.

---

We use the Binomial Theorem for expansion:

$(a+b)^n = \binom{n}{0}a^n b^0 + \binom{n}{1}a^{n-1} b^1 + \binom{n}{2}a^{n-2} b^2 + \binom{n}{3}a^{n-3} b^3 + \binom{n}{4}a^{n-4} b^4$

For $n=4$, the expansion of $(100 + 1)^4$ is:

$(100 + 1)^4 = \binom{4}{0}(100)^4 (1)^0 + \binom{4}{1}(100)^3 (1)^1 + \binom{4}{2}(100)^2 (1)^2 + \binom{4}{3}(100)^1 (1)^3 + \binom{4}{4}(100)^0 (1)^4$

---

Now, we calculate each term:

Term 1: $\binom{4}{0}(100)^4 (1)^0 = 1 \times 10^8 \times 1 = 100,000,000$

Term 2: $\binom{4}{1}(100)^3 (1)^1 = 4 \times 10^6 \times 1 = 4 \times 1,000,000 = 4,000,000$

Term 3: $\binom{4}{2}(100)^2 (1)^2 = 6 \times 10^4 \times 1 = 6 \times 10,000 = 60,000$

Term 4: $\binom{4}{3}(100)^1 (1)^3 = 4 \times 100 \times 1 = 400$

Term 5: $\binom{4}{4}(100)^0 (1)^4 = 1 \times 1 \times 1 = 1$

---

Adding the terms together:

$(101)^4 = 100,000,000 + 4,000,000 + 60,000 + 400 + 1$

$(101)^4 = 104,000,000 + 60,000 + 400 + 1$

$(101)^4 = 104,060,000 + 400 + 1$

$(101)^4 = 104,060,400 + 1$

$(101)^4 = 104,060,401$

---

The evaluated value is:

$(101)^4 = \mathbf{104,060,401}$

Given expression:

$(99)^5$

---

We can write 99 as $100 - 1$. So, we need to evaluate $(100 - 1)^5$.

This is in the form $(a-b)^n$, where $a = 100$, $b = 1$, and $n = 5$.

---

We use the Binomial Theorem for expansion:

$(a-b)^n = \binom{n}{0}a^n b^0 - \binom{n}{1}a^{n-1} b^1 + \binom{n}{2}a^{n-2} b^2 - \binom{n}{3}a^{n-3} b^3 + \binom{n}{4}a^{n-4} b^4 - \binom{n}{5}a^{n-5} b^5$

For $n=5$, the expansion of $(100 - 1)^5$ is:

$(100 - 1)^5 = \binom{5}{0}(100)^5 (1)^0 - \binom{5}{1}(100)^4 (1)^1 + \binom{5}{2}(100)^3 (1)^2 - \binom{5}{3}(100)^2 (1)^3 + \binom{5}{4}(100)^1 (1)^4 - \binom{5}{5}(100)^0 (1)^5$

---

Now, we calculate each term:

Term 1: $\binom{5}{0}(100)^5 (1)^0 = 1 \times 10^{10} \times 1 = 10,000,000,000$

Term 2: $-\binom{5}{1}(100)^4 (1)^1 = -5 \times 10^8 \times 1 = -5 \times 100,000,000 = -500,000,000$

Term 3: $\binom{5}{2}(100)^3 (1)^2 = 10 \times 10^6 \times 1 = 10 \times 1,000,000 = 10,000,000$

Term 4: $-\binom{5}{3}(100)^2 (1)^3 = -10 \times 10^4 \times 1 = -10 \times 10,000 = -100,000$

Term 5: $\binom{5}{4}(100)^1 (1)^4 = 5 \times 100 \times 1 = 500$

Term 6: $-\binom{5}{5}(100)^0 (1)^5 = -1 \times 1 \times 1 = -1$

---

Adding the terms together:

$(99)^5 = 10,000,000,000 - 500,000,000 + 10,000,000 - 100,000 + 500 - 1$

$(99)^5 = 9,500,000,000 + 10,000,000 - 100,000 + 500 - 1$

$(99)^5 = 9,510,000,000 - 100,000 + 500 - 1$

$(99)^5 = 9,509,900,000 + 500 - 1$

$(99)^5 = 9,509,900,500 - 1$

$(99)^5 = 9,509,900,499$

---

The evaluated value is:

$(99)^5 = \mathbf{9,509,900,499}$

We need to compare the values of $(1.1)^{10000}$ and $1000$.

---

Consider the expression $(1.1)^{10000}$.

We can write $1.1$ as $1 + 0.1$.

So, $(1.1)^{10000} = (1 + 0.1)^{10000}$.

---

We can expand this expression using the Binomial Theorem for $(1+x)^n$, which is given by:

$(1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \binom{n}{3}x^3 + ... + \binom{n}{n}x^n$

In this case, $x = 0.1$ and $n = 10000$.

Expanding $(1 + 0.1)^{10000}$:

$(1 + 0.1)^{10000} = \binom{10000}{0} + \binom{10000}{1}(0.1) + \binom{10000}{2}(0.1)^2 + \binom{10000}{3}(0.1)^3 + ... + \binom{10000}{10000}(0.1)^{10000}$

---

Let's calculate the first two terms of the expansion:

The first term is $\binom{10000}{0} = 1$.

The second term is $\binom{10000}{1}(0.1) = 10000 \times 0.1 = 1000$.

---

So, the expansion can be written as:

$(1.1)^{10000} = 1 + 1000 + \binom{10000}{2}(0.1)^2 + \binom{10000}{3}(0.1)^3 + ... + \binom{10000}{10000}(0.1)^{10000}$

$(1.1)^{10000} = 1001 + \binom{10000}{2}(0.1)^2 + \binom{10000}{3}(0.1)^3 + ... + \binom{10000}{10000}(0.1)^{10000}$

---

The terms $\binom{10000}{k}(0.1)^k$ for $k \geq 2$ are all positive because $\binom{10000}{k}$ is positive for $k \geq 0$, and $(0.1)^k$ is positive for $k \geq 2$.

The sum of these positive terms is a positive value.

Let $S = \binom{10000}{2}(0.1)^2 + \binom{10000}{3}(0.1)^3 + ... + \binom{10000}{10000}(0.1)^{10000}$. Here $S > 0$.

So, $(1.1)^{10000} = 1001 + S$.

---

Since $S$ is a positive value, $1001 + S$ is greater than $1001$.

$(1.1)^{10000} > 1001$

We are comparing $(1.1)^{10000}$ with $1000$.

Since $(1.1)^{10000} > 1001$ and $1001 > 1000$, by transitivity, we have $(1.1)^{10000} > 1000$.

---

Therefore, using the Binomial Theorem, we can conclude that $\mathbf{(1.1)^{10000}}$ is larger than $1000$.

First, let's find the expansion of $(a+b)^4$ and $(a-b)^4$ using the Binomial Theorem.

---

The expansion of $(a+b)^4$ is:

$(a+b)^4 = \binom{4}{0}a^4 b^0 + \binom{4}{1}a^3 b^1 + \binom{4}{2}a^2 b^2 + \binom{4}{3}a^1 b^3 + \binom{4}{4}a^0 b^4$

$(a+b)^4 = 1 \cdot a^4 \cdot 1 + 4 \cdot a^3 b + 6 \cdot a^2 b^2 + 4 \cdot a b^3 + 1 \cdot 1 \cdot b^4$

$(a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$

---

The expansion of $(a-b)^4$ is:

$(a-b)^4 = \binom{4}{0}a^4 (-b)^0 + \binom{4}{1}a^3 (-b)^1 + \binom{4}{2}a^2 (-b)^2 + \binom{4}{3}a^1 (-b)^3 + \binom{4}{4}a^0 (-b)^4$

$(a-b)^4 = 1 \cdot a^4 \cdot 1 + 4 \cdot a^3 (-b) + 6 \cdot a^2 b^2 + 4 \cdot a (-b^3) + 1 \cdot 1 \cdot b^4$

$(a-b)^4 = a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4$

---

Now, we subtract $(a-b)^4$ from $(a+b)^4$:

$(a+b)^4 - (a-b)^4 = (a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4) - (a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4)$

$(a+b)^4 - (a-b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 - a^4 + 4a^3b - 6a^2b^2 + 4ab^3 - b^4$

Combine like terms:

$(a+b)^4 - (a-b)^4 = (a^4 - a^4) + (4a^3b + 4a^3b) + (6a^2b^2 - 6a^2b^2) + (4ab^3 + 4ab^3) + (b^4 - b^4)$

$(a+b)^4 - (a-b)^4 = 0 + 8a^3b + 0 + 8ab^3 + 0$

$(a+b)^4 - (a-b)^4 = 8a^3b + 8ab^3$

Factor out $8ab$:

$(a+b)^4 - (a-b)^4 = 8ab(a^2 + b^2)$

---

Now, we need to evaluate $(\sqrt{3} + \sqrt{2})^4 - (\sqrt{3} - \sqrt{2})^4$.

This expression is in the form $(a+b)^4 - (a-b)^4$, where $a = \sqrt{3}$ and $b = \sqrt{2}$.

Using the derived formula $8ab(a^2 + b^2)$:

Substitute $a = \sqrt{3}$ and $b = \sqrt{2}$:

$8(\sqrt{3})(\sqrt{2})((\sqrt{3})^2 + (\sqrt{2})^2)$

Calculate the terms inside the expression:

$(\sqrt{3})(\sqrt{2}) = \sqrt{3 \times 2} = \sqrt{6}$

$(\sqrt{3})^2 = 3$

$(\sqrt{2})^2 = 2$

$(\sqrt{3})^2 + (\sqrt{2})^2 = 3 + 2 = 5$

---

Substitute these values back into the formula:

$8(\sqrt{6})(5)$

$8 \times 5 \times \sqrt{6} = 40\sqrt{6}$

---

The value of $(\sqrt{3} + \sqrt{2})^4 - (\sqrt{3} - \sqrt{2})^4$ is $\mathbf{40\sqrt{6}}$.

First, let's find the expansion of $(x+1)^6$ and $(x-1)^6$ using the Binomial Theorem.

---

The expansion of $(x+1)^6$ is of the form $(a+b)^n$ with $a=x$, $b=1$, $n=6$.

$(x+1)^6 = \binom{6}{0}x^6 1^0 + \binom{6}{1}x^5 1^1 + \binom{6}{2}x^4 1^2 + \binom{6}{3}x^3 1^3 + \binom{6}{4}x^2 1^4 + \binom{6}{5}x^1 1^5 + \binom{6}{6}x^0 1^6$

$(x+1)^6 = 1 \cdot x^6 + 6 \cdot x^5 \cdot 1 + 15 \cdot x^4 \cdot 1 + 20 \cdot x^3 \cdot 1 + 15 \cdot x^2 \cdot 1 + 6 \cdot x \cdot 1 + 1 \cdot 1 \cdot 1$

$(x+1)^6 = x^6 + 6x^5 + 15x^4 + 20x^3 + 15x^2 + 6x + 1$

---

The expansion of $(x-1)^6$ is of the form $(a-b)^n$ with $a=x$, $b=1$, $n=6$. The signs of terms with odd powers of the second term ($(-1)^k$) will be negative.

$(x-1)^6 = \binom{6}{0}x^6 (-1)^0 + \binom{6}{1}x^5 (-1)^1 + \binom{6}{2}x^4 (-1)^2 + \binom{6}{3}x^3 (-1)^3 + \binom{6}{4}x^2 (-1)^4 + \binom{6}{5}x^1 (-1)^5 + \binom{6}{6}x^0 (-1)^6$

$(x-1)^6 = 1 \cdot x^6 \cdot 1 + 6 \cdot x^5 \cdot (-1) + 15 \cdot x^4 \cdot 1 + 20 \cdot x^3 \cdot (-1) + 15 \cdot x^2 \cdot 1 + 6 \cdot x \cdot (-1) + 1 \cdot 1 \cdot 1$

$(x-1)^6 = x^6 - 6x^5 + 15x^4 - 20x^3 + 15x^2 - 6x + 1$

---

Now, we add $(x+1)^6$ and $(x-1)^6$:

$(x+1)^6 + (x-1)^6 = (x^6 + 6x^5 + 15x^4 + 20x^3 + 15x^2 + 6x + 1) + (x^6 - 6x^5 + 15x^4 - 20x^3 + 15x^2 - 6x + 1)$

Combine like terms:

$(x+1)^6 + (x-1)^6 = (x^6 + x^6) + (6x^5 - 6x^5) + (15x^4 + 15x^4) + (20x^3 - 20x^3) + (15x^2 + 15x^2) + (6x - 6x) + (1 + 1)$

$(x+1)^6 + (x-1)^6 = 2x^6 + 0 + 30x^4 + 0 + 30x^2 + 0 + 2$

$(x+1)^6 + (x-1)^6 = 2x^6 + 30x^4 + 30x^2 + 2$

Factor out 2:

$(x+1)^6 + (x-1)^6 = 2(x^6 + 15x^4 + 15x^2 + 1)$

---

Now, we need to evaluate $(\sqrt{2} + 1)^6 + (\sqrt{2} - 1)^6$.

This expression is in the form $(x+1)^6 + (x-1)^6$, where $x = \sqrt{2}$.

Using the derived formula $2(x^6 + 15x^4 + 15x^2 + 1)$:

Substitute $x = \sqrt{2}$:

$2((\sqrt{2})^6 + 15(\sqrt{2})^4 + 15(\sqrt{2})^2 + 1)$

Calculate the powers of $\sqrt{2}$:

$(\sqrt{2})^2 = 2$

$(\sqrt{2})^4 = ((\sqrt{2})^2)^2 = 2^2 = 4$

$(\sqrt{2})^6 = ((\sqrt{2})^2)^3 = 2^3 = 8$

---

Substitute these values back into the expression:

$2(8 + 15(4) + 15(2) + 1)$

$2(8 + 60 + 30 + 1)$

$2(68 + 30 + 1)$

$2(98 + 1)$

$2(99)$

$2 \times 99 = 198$

---

The value of $(\sqrt{2} + 1)^6 + (\sqrt{2} - 1)^6$ is $\mathbf{198}$.

To Show:

The expression $9^{n+1} - 8n - 9$ is divisible by 64 for any positive integer $n$.

This is equivalent to showing that $9^{n+1} - 8n - 9 \equiv 0 \pmod{64}$, or $9^{n+1} - 8n - 9 = 64K$ for some integer $K$.

---

Proof:

Consider the term $9^{n+1}$. We can write $9$ as $1 + 8$.

So, $9^{n+1} = (1 + 8)^{n+1}$.

Using the Binomial Theorem for a positive integer $n$, the expansion of $(1+x)^m$ is given by:

$(1+x)^m = \binom{m}{0} + \binom{m}{1}x + \binom{m}{2}x^2 + \binom{m}{3}x^3 + ... + \binom{m}{m}x^m$

Substitute $x = 8$ and $m = n+1$ into the binomial expansion of $(1+8)^{n+1}$:

$9^{n+1} = (1+8)^{n+1} = \binom{n+1}{0}1^{n+1} 8^0 + \binom{n+1}{1}1^n 8^1 + \binom{n+1}{2}1^{n-1} 8^2 + \binom{n+1}{3}1^{n-2} 8^3 + ... + \binom{n+1}{n+1}1^0 8^{n+1}$

$9^{n+1} = \binom{n+1}{0} + \binom{n+1}{1}8 + \binom{n+1}{2}8^2 + \binom{n+1}{3}8^3 + ... + \binom{n+1}{n+1}8^{n+1}$

---

We know that $\binom{n+1}{0} = 1$ and $\binom{n+1}{1} = n+1$. Also, $8^2 = 64$, $8^3 = 512$, and so on.

So, the expansion becomes:

$9^{n+1} = 1 + (n+1) \times 8 + \binom{n+1}{2} \times 64 + \binom{n+1}{3} \times 512 + ... + \binom{n+1}{n+1} \times 8^{n+1}$

$9^{n+1} = 1 + 8(n+1) + 64\binom{n+1}{2} + 512\binom{n+1}{3} + ... + 8^{n+1}\binom{n+1}{n+1}$

$9^{n+1} = 1 + 8n + 8 + 64\binom{n+1}{2} + 512\binom{n+1}{3} + ... + 8^{n+1}\binom{n+1}{n+1}$

$9^{n+1} = 9 + 8n + 64\binom{n+1}{2} + 512\binom{n+1}{3} + ... + 8^{n+1}\binom{n+1}{n+1}$

---

Now, consider the expression $9^{n+1} - 8n - 9$. Substitute the expansion of $9^{n+1}$ into this expression:

$9^{n+1} - 8n - 9 = (9 + 8n + 64\binom{n+1}{2} + 512\binom{n+1}{3} + ... + 8^{n+1}\binom{n+1}{n+1}) - 8n - 9$

$9^{n+1} - 8n - 9 = (9 - 9) + (8n - 8n) + 64\binom{n+1}{2} + 512\binom{n+1}{3} + ... + 8^{n+1}\binom{n+1}{n+1}$

$9^{n+1} - 8n - 9 = 0 + 0 + 64\binom{n+1}{2} + 512\binom{n+1}{3} + ... + 8^{n+1}\binom{n+1}{n+1}$

$9^{n+1} - 8n - 9 = 64\binom{n+1}{2} + 512\binom{n+1}{3} + ... + 8^{n+1}\binom{n+1}{n+1}$

---

For $n=1$: The expression is $9^2 - 8(1) - 9 = 81 - 8 - 9 = 64$. $64$ is divisible by $64$.

Using the formula for $n=1$: $9^{1+1} - 8(1) - 9 = 64\binom{1+1}{2} + 512\binom{1+1}{3} + ... = 64\binom{2}{2} + 512\binom{2}{3} + ... = 64(1) + 512(0) + ... = 64$. This confirms the formula holds for $n=1$. For $n=1$, the expansion stops at the term with $8^2 = 64$.

---

For $n \geq 1$, every term $8^k \binom{n+1}{k}$ for $k \geq 2$ in the expansion of $9^{n+1} - 8n - 9$ has a factor of $8^2 = 64$ because $8^k = 8^{k-2} \times 8^2 = 8^{k-2} \times 64$.

We can factor out 64 from all the terms:

$9^{n+1} - 8n - 9 = 64\binom{n+1}{2} + 64 \times 8\binom{n+1}{3} + ... + 64 \times 8^{n-1}\binom{n+1}{n+1}$

$9^{n+1} - 8n - 9 = 64\left[\binom{n+1}{2} + 8\binom{n+1}{3} + ... + 8^{n-1}\binom{n+1}{n+1}\right]$

Let $K = \binom{n+1}{2} + 8\binom{n+1}{3} + ... + 8^{n-1}\binom{n+1}{n+1}$. Since $n$ is a positive integer, $\binom{n+1}{k}$ are integers, and powers of 8 are integers, $K$ is an integer for $n \geq 1$.

So, for any positive integer $n$, we can write:

$9^{n+1} - 8n - 9 = 64K$

where $K$ is an integer.

---

This shows that $9^{n+1} - 8n - 9$ is a multiple of 64 for any positive integer $n$.

Therefore, $9^{n+1} - 8n - 9$ is divisible by 64 whenever $n$ is a positive integer.

Hence Showed.

To Prove:

$\sum\limits_{r=0}^{n} 3^r\; ^nC_r = 4^n$

---

Proof:

We start with the statement of the Binomial Theorem for $(a+b)^n$:

$(a+b)^n = \sum\limits_{r=0}^{n} \binom{n}{r} a^{n-r} b^r$

We can rewrite $\binom{n}{r}$ as $^nC_r$. So, the Binomial Theorem is:

$(a+b)^n = \sum\limits_{r=0}^{n} ^nC_r a^{n-r} b^r$

---

We want the summation to match the form $\sum\limits_{r=0}^{n} 3^r\; ^nC_r$.

Comparing the term $^nC_r 3^r$ with the general term $^nC_r a^{n-r} b^r$, we can see that the exponent of $a$ in the sum is 0 ($n-r = n-r$) and the exponent of $b$ is $r$.

This suggests we can choose the values of $a$ and $b$ appropriately.

If we choose $b = 3$, the term becomes $^nC_r a^{n-r} 3^r$.

To get the desired form $^nC_r 3^r$, we need the term $a^{n-r}$ to be 1 for all $r$ from 0 to $n$. This can be achieved by setting $a=1$, since $1^{n-r} = 1$ for any non-negative integer $n-r$.

---

Let's substitute $a=1$ and $b=3$ into the Binomial Theorem:

$(1+3)^n = \sum\limits_{r=0}^{n} ^nC_r (1)^{n-r} (3)^r$

$(1+3)^n = \sum\limits_{r=0}^{n} ^nC_r \times 1 \times 3^r$

$(1+3)^n = \sum\limits_{r=0}^{n} ^nC_r 3^r$

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The left side of the equation is $(1+3)^n = 4^n$.

The right side of the equation is $\sum\limits_{r=0}^{n} ^nC_r 3^r$, which is the same as $\sum\limits_{r=0}^{n} 3^r\; ^nC_r$.

Therefore, we have shown that:

$4^n = \sum\limits_{r=0}^{n} 3^r\; ^nC_r$

Or, written in the requested format:

$\sum\limits_{r=0}^{n} 3^r\; ^nC_r = 4^n$

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Hence Proved.

Given expansion:

$(2 + a)^{50}$

This is in the form $(x+y)^n$, where $x = 2$, $y = a$, and $n = 50$.

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The general term (the $(r+1)^{\text{th}}$ term) in the expansion of $(x+y)^n$ is given by the formula:

$T_{r+1} = \binom{n}{r} x^{n-r} y^r$

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We are given that the 17^th term and the 18^th term of the expansion are equal.

The 17^th term corresponds to $r+1 = 17$, so $r = 16$.

$T_{17} = T_{16+1} = \binom{50}{16} (2)^{50-16} (a)^{16} = \binom{50}{16} 2^{34} a^{16}$

The 18^th term corresponds to $r+1 = 18$, so $r = 17$.

$T_{18} = T_{17+1} = \binom{50}{17} (2)^{50-17} (a)^{17} = \binom{50}{17} 2^{33} a^{17}$

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We are given that $T_{17} = T_{18}$.

$\binom{50}{16} 2^{34} a^{16} = \binom{50}{17} 2^{33} a^{17}$

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We know the formula for binomial coefficients: $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

$\binom{50}{16} = \frac{50!}{16!(50-16)!} = \frac{50!}{16!34!}$

$\binom{50}{17} = \frac{50!}{17!(50-17)!} = \frac{50!}{17!33!}$

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Substitute these into the equation:

$\frac{50!}{16!34!} \times 2^{34} \times a^{16} = \frac{50!}{17!33!} \times 2^{33} \times a^{17}$

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We can simplify the factorials: $17! = 17 \times 16!$ and $34! = 34 \times 33!$.

$\frac{50!}{16! (34 \times 33!)} \times 2^{34} \times a^{16} = \frac{50!}{(17 \times 16!) 33!} \times 2^{33} \times a^{17}$

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Cancel out the common terms $\frac{50!}{16!33!}$ from both sides (assuming $a \neq 0$ and the terms are non-zero):

$\frac{1}{34} \times 2^{34} \times a^{16} = \frac{1}{17} \times 2^{33} \times a^{17}$

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Also, $2^{34} = 2 \times 2^{33}$ and $a^{17} = a \times a^{16}$.

$\frac{1}{34} \times (2 \times 2^{33}) \times a^{16} = \frac{1}{17} \times 2^{33} \times (a \times a^{16})$

$\frac{2}{34} \times 2^{33} \times a^{16} = \frac{1}{17} \times 2^{33} \times a^{16} \times a$

$\frac{1}{17} \times 2^{33} \times a^{16} = \frac{1}{17} \times 2^{33} \times a^{16} \times a$

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If $2^{33} a^{16} \neq 0$, we can divide both sides by $\frac{1}{17} \times 2^{33} \times a^{16}$.

Note that $a=0$ would make $T_{18}=0$, but $T_{17}$ would be $\binom{50}{16} 2^{34} \neq 0$, so $T_{17} \neq T_{18}$. Thus $a \neq 0$.

Dividing by $\frac{1}{17} \times 2^{33} \times a^{16}$:

$1 = a$

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Alternatively, from $\frac{1}{34} \times 2^{34} \times a^{16} = \frac{1}{17} \times 2^{33} \times a^{17}$, rearrange to solve for $a$ (assuming $a \neq 0$):

$\frac{a^{17}}{a^{16}} = \frac{\frac{1}{34} \times 2^{34}}{\frac{1}{17} \times 2^{33}}$

$a = \frac{1}{34} \times 2^{34} \times \frac{17}{1} \times \frac{1}{2^{33}}$

$a = \frac{17}{34} \times \frac{2^{34}}{2^{33}}$

$a = \frac{1}{2} \times 2^{34-33}$

$a = \frac{1}{2} \times 2^1$

$a = \frac{1}{2} \times 2 = 1$

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The value of $a$ is 1.

Given expansion:

$(1 + x)^{2n}$

This is in the form $(a+b)^m$, where $a = 1$, $b = x$, and $m = 2n$.

The total number of terms in the expansion is $m+1 = 2n+1$.

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To Find:

The middle term of the expansion.

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Solution:

Since the number of terms ($2n+1$) is odd, there is exactly one middle term.

The position of the middle term is $\left(\frac{(2n+1)+1}{2}\right)^{\text{th}}$ term.

Middle term position = $\left(\frac{2n+2}{2}\right)^{\text{th}}$ term = $(n+1)^{\text{th}}$ term.

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The general term (the $(r+1)^{\text{th}}$ term) in the expansion of $(a+b)^m$ is given by the formula:

$T_{r+1} = \binom{m}{r} a^{m-r} b^r$

For the $(n+1)^{\text{th}}$ term, we have $r+1 = n+1$, so $r = n$.

Substitute $m=2n$, $a=1$, $b=x$, and $r=n$ into the general term formula:

$T_{n+1} = \binom{2n}{n} (1)^{2n-n} (x)^n = \binom{2n}{n} (1)^n x^n = \binom{2n}{n} x^n$

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Now, we need to express the binomial coefficient $\binom{2n}{n}$ in the desired form.

$\binom{2n}{n} = \frac{(2n)!}{n!(2n-n)!} = \frac{(2n)!}{n!n!}$

Expand $(2n)!$:

$(2n)! = (2n) \times (2n-1) \times (2n-2) \times (2n-3) \times ... \times 4 \times 3 \times 2 \times 1$

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We can separate the even and odd factors in $(2n)!$:

$(2n)! = [(2n) \times (2n-2) \times (2n-4) \times ... \times 4 \times 2] \times [(2n-1) \times (2n-3) \times ... \times 3 \times 1]$

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The first bracket contains $n$ even terms. We can factor out 2 from each even term:

$(2n) \times (2n-2) \times ... \times 4 \times 2 = (2 \times n) \times (2 \times (n-1)) \times ... \times (2 \times 2) \times (2 \times 1)$

$= 2^n \times [n \times (n-1) \times ... \times 2 \times 1] = 2^n \times n!$

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The second bracket contains the product of $n$ odd numbers:

$(2n-1) \times (2n-3) \times ... \times 3 \times 1 = 1 \times 3 \times 5 \times ... \times (2n-1)$

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So, $(2n)! = (2^n \times n!) \times (1 \times 3 \times 5 \times ... \times (2n-1))$

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Now substitute this back into the expression for $\binom{2n}{n}$:

$\binom{2n}{n} = \frac{(2n)!}{n!n!} = \frac{(2^n \times n!) \times (1 \times 3 \times 5 \times ... \times (2n-1))}{n!n!}$

Cancel one $n!$ from the numerator and denominator:

$\binom{2n}{n} = \frac{2^n \times (1 \times 3 \times 5 \times ... \times (2n-1))}{n!}$

$\binom{2n}{n} = \frac{1 \cdot 3 \cdot 5 \cdot ... \cdot (2n-1)}{n!} \times 2^n$

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Substitute this expression for $\binom{2n}{n}$ back into the middle term $T_{n+1} = \binom{2n}{n} x^n$:

$T_{n+1} = \left(\frac{1 \cdot 3 \cdot 5 \cdot ... \cdot (2n-1)}{n!} \times 2^n\right) x^n$

$T_{n+1} = \frac{1 \cdot 3 \cdot 5 \cdot ... \cdot (2n-1)}{n!} 2^n x^n$

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This matches the form given in the question, except for the factor of 2n, which seems to be a typo in the question (it should be $2^n$). Let's double check the calculation and the requested form.

The requested form is $\frac{1.3.5...(2n \;-\; 1)}{n!}$ 2n xⁿ. There seems to be a discrepancy with the factor of 2n vs $2^n$. Let's assume the question meant $2^n$. If the question meant 2n as a coefficient, it would have to come from somewhere else, which isn't apparent from the expansion.

Let's proceed assuming the question meant $2^n$.

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The middle term is $T_{n+1} = \binom{2n}{n} x^n$. We have shown that $\binom{2n}{n} = \frac{(2n)!}{(n!)^2} = \frac{1 \cdot 3 \cdot 5 \cdot ... \cdot (2n-1) \cdot 2^n \cdot n!}{n! \cdot n!} = \frac{1 \cdot 3 \cdot 5 \cdot ... \cdot (2n-1) \cdot 2^n}{n!}$.

So, $T_{n+1} = \frac{1 \cdot 3 \cdot 5 \cdot ... \cdot (2n-1) \cdot 2^n}{n!} x^n$.

This is exactly the form $\frac{1.3.5...(2n \;-\; 1)}{n!} 2^n x^n$.

If the question specifically meant 2n instead of $2^n$, there might be a different approach or a typo in the question itself.

Assuming the question has a typo and meant $2^n$ instead of 2n, the proof is complete.

If the question is exactly as written, it is likely incorrect or requires manipulation that is not standard Binomial Theorem expansion.

Let's re-read the question carefully. "Show that the middle term ... is $\frac{1.3.5...(2n \;-\; 1)}{n!}$ 2n xⁿ". Yes, it explicitly says 2n.

Let's re-examine the binomial coefficient $\binom{2n}{n} = \frac{(2n)!}{n!n!}$.

$\binom{2n}{n} = \frac{1 \cdot 2 \cdot 3 \cdot ... \cdot (2n-1) \cdot (2n)}{n! n!}$

Maybe the 2n comes from somewhere else?

Let's look at the structure $\frac{1.3.5...(2n \;-\; 1)}{n!} \times 2n \times x^n$.

This equals $\frac{1 \cdot 3 \cdot 5 \cdot ... \cdot (2n-1)}{n!} \times 2n \times x^n$.

The binomial coefficient is $\frac{(2n)!}{(n!)^2} = \frac{1 \cdot 3 \cdot ... \cdot (2n-1) \cdot 2 \cdot 4 \cdot ... \cdot (2n)}{n! n!} = \frac{1 \cdot 3 \cdot ... \cdot (2n-1) \cdot 2^n \cdot n!}{n! n!} = \frac{1 \cdot 3 \cdot ... \cdot (2n-1) \cdot 2^n}{n!}$.

So the middle term is $\frac{1 \cdot 3 \cdot 5 \cdot ... \cdot (2n-1) \cdot 2^n}{n!} x^n$.

There is a factor of $2^n$ in the correct expansion, not $2n$. It is highly probable that there is a typo in the question statement and 2n should be $2^n$.

Assuming the question meant $2^n$, the proof above is correct.

Let's write the answer based on the assumption of the typo.

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The middle term in the expansion of $(1+x)^{2n}$ is the $(n+1)^{\text{th}}$ term.

$T_{n+1} = \binom{2n}{n} 1^{2n-n} x^n = \binom{2n}{n} x^n$

We have $\binom{2n}{n} = \frac{(2n)!}{n!n!}$

$(2n)! = 1 \cdot 2 \cdot 3 \cdot 4 \cdot ... \cdot (2n-1) \cdot (2n)$

$(2n)! = [1 \cdot 3 \cdot 5 \cdot ... \cdot (2n-1)] \times [2 \cdot 4 \cdot 6 \cdot ... \cdot (2n)]$

$(2n)! = [1 \cdot 3 \cdot 5 \cdot ... \cdot (2n-1)] \times [2 \cdot (2 \times 2) \cdot (2 \times 3) \cdot ... \cdot (2 \times n)]$

$(2n)! = [1 \cdot 3 \cdot 5 \cdot ... \cdot (2n-1)] \times [2^n \cdot (1 \cdot 2 \cdot 3 \cdot ... \cdot n)]$

$(2n)! = [1 \cdot 3 \cdot 5 \cdot ... \cdot (2n-1)] \times 2^n \times n!$

So, $\binom{2n}{n} = \frac{[1 \cdot 3 \cdot 5 \cdot ... \cdot (2n-1)] \times 2^n \times n!}{n!n!} = \frac{1 \cdot 3 \cdot 5 \cdot ... \cdot (2n-1) \times 2^n}{n!}$

The middle term is $T_{n+1} = \binom{2n}{n} x^n = \frac{1 \cdot 3 \cdot 5 \cdot ... \cdot (2n-1) \times 2^n}{n!} x^n$

We can write this as $\frac{1 \cdot 3 \cdot 5 \cdot ... \cdot (2n-1)}{n!} 2^n x^n$.

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If the question indeed meant 2n instead of $2^n$, then the statement is incorrect. However, standard results show the factor is $2^n$. Given this is an example question, it is very likely a typo in the textbook/source of the question.

Assuming the question intended to have $2^n$ instead of $2n$, we have proved the statement.

The middle term in the expansion of $(1 + x)^{2n}$ is $\mathbf{\frac{1.3.5...(2n \;-\; 1)}{n!} 2^n x^n}$.

Note: There is likely a typo in the question, where 2n should be $2^n$. The derived formula matches the standard result with $2^n$. Assuming the standard result is what needs to be shown, the proof holds.

Given expansion:

$(x + 2y)^9$

This is in the form $(a+b)^n$, where $a = x$, $b = 2y$, and $n = 9$.

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To Find:

The coefficient of the term $x^6y^3$ in the expansion.

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Solution:

The general term (the $(r+1)^{\text{th}}$ term) in the expansion of $(a+b)^n$ is given by the formula:

$T_{r+1} = \binom{n}{r} a^{n-r} b^r$

Substitute the values $n=9$, $a=x$, and $b=2y$:

$T_{r+1} = \binom{9}{r} (x)^{9-r} (2y)^r$

$T_{r+1} = \binom{9}{r} x^{9-r} 2^r y^r$

$T_{r+1} = \binom{9}{r} 2^r x^{9-r} y^r$

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We are looking for the term containing $x^6y^3$.

Comparing the exponents of $x$ and $y$ in the general term with those in $x^6y^3$:

For the exponent of $x$: $9-r = 6 \implies r = 9 - 6 = 3$.

For the exponent of $y$: $r = 3$.

Both conditions give $r = 3$. So, the term we are interested in is the $(3+1)^{\text{th}}$, which is the 4th term.

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Substitute $r = 3$ into the general term formula to find the term:

$T_{3+1} = \binom{9}{3} 2^3 x^{9-3} y^3$

$T_4 = \binom{9}{3} 2^3 x^6 y^3$

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Now, calculate the value of the coefficient, which is $\binom{9}{3} 2^3$.

$\binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!} = \frac{9 \times 8 \times 7 \times 6!}{3 \times 2 \times 1 \times 6!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = \frac{504}{6} = 84$

$2^3 = 2 \times 2 \times 2 = 8$

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The coefficient of $x^6y^3$ is $\binom{9}{3} 2^3 = 84 \times 8$.

Calculation:

$84 \times 8 = 672$

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The term containing $x^6y^3$ is $672 x^6y^3$.

The coefficient of $x^6y^3$ is $\mathbf{672}$.

Given expansion:

$(x + a)^n$

The terms are in the form $(a+b)^n$, with $a$ replaced by $x$ and $b$ replaced by $a$. Let's use $u$ and $v$ for the terms in the binomial expansion to avoid confusion with the variable $a$ in the question. The expansion is $(u+v)^n$ where $u=x$ and $v=a$.

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The general term (the $(r+1)^{\text{th}}$ term) in the expansion of $(u+v)^n$ is given by:

$T_{r+1} = \binom{n}{r} u^{n-r} v^r$

Substituting $u=x$ and $v=a$:

$T_{r+1} = \binom{n}{r} x^{n-r} a^r$

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We are given the values of the second, third, and fourth terms:

Second term, $T_2$ (where $r+1=2$, so $r=1$):

$T_2 = \binom{n}{1} x^{n-1} a^1 = nx^{n-1}a = 240$

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Third term, $T_3$ (where $r+1=3$, so $r=2$):

$T_3 = \binom{n}{2} x^{n-2} a^2 = \frac{n(n-1)}{2}x^{n-2}a^2 = 720$

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Fourth term, $T_4$ (where $r+1=4$, so $r=3$):

$T_4 = \binom{n}{3} x^{n-3} a^3 = \frac{n(n-1)(n-2)}{3 \times 2 \times 1}x^{n-3}a^3 = \frac{n(n-1)(n-2)}{6}x^{n-3}a^3 = 1080$

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To find $n$ and $\frac{a}{x}$, divide the third term by the second term:

$\frac{T_3}{T_2} = \frac{\frac{n(n-1)}{2}x^{n-2}a^2}{nx^{n-1}a} = \frac{720}{240}$

$\frac{n(n-1)}{2n} \frac{x^{n-2}}{x^{n-1}} \frac{a^2}{a} = 3$

$\frac{n-1}{2} x^{(n-2)-(n-1)} a^{2-1} = 3$

$\frac{n-1}{2} x^{-1} a^1 = 3$

$\frac{n-1}{2} \frac{a}{x} = 3$

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Now, divide the fourth term by the third term:

$\frac{T_4}{T_3} = \frac{\frac{n(n-1)(n-2)}{6}x^{n-3}a^3}{\frac{n(n-1)}{2}x^{n-2}a^2} = \frac{1080}{720}$

$\frac{n(n-1)(n-2)}{6} \times \frac{2}{n(n-1)} \times \frac{x^{n-3}}{x^{n-2}} \times \frac{a^3}{a^2} = \frac{108}{72} = \frac{54}{36} = \frac{3}{2}$

$\frac{n-2}{3} x^{(n-3)-(n-2)} a^{3-2} = \frac{3}{2}$

$\frac{n-2}{3} x^{-1} a^1 = \frac{3}{2}$

$\frac{n-2}{3} \frac{a}{x} = \frac{3}{2}$

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We now have a system of two equations with two unknowns, $n$ and $\frac{a}{x}$.

Divide equation (iv) by equation (v):

$\frac{(n-1)\frac{a}{x}}{(n-2)\frac{a}{x}} = \frac{6}{\frac{9}{2}}$

$\frac{n-1}{n-2} = 6 \times \frac{2}{9} = \frac{12}{9} = \frac{4}{3}$

Cross-multiply:

$3(n-1) = 4(n-2)$

$3n - 3 = 4n - 8$

$8 - 3 = 4n - 3n$

$5 = n$

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Now that we have $n=5$, substitute this value into equation (iv) to find $\frac{a}{x}$:

$(5-1)\frac{a}{x} = 6$

$4\frac{a}{x} = 6$

$\frac{a}{x} = \frac{6}{4} = \frac{3}{2}$

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Finally, substitute $n=5$ into equation (i) to find the relationship between $x$ and $a$, and then use the value of $\frac{a}{x}$ to solve for $x$ and $a$ individually.

Equation (i): $nx^{n-1}a = 240$

$5x^{5-1}a = 240$

$5x^4a = 240$

$x^4a = \frac{240}{5} = 48$

From equation (vii), $a = \frac{3}{2}x$. Substitute this into equation (viii):

$x^4 \left(\frac{3}{2}x\right) = 48$

$\frac{3}{2} x^{4+1} = 48$

$\frac{3}{2} x^5 = 48$

$x^5 = 48 \times \frac{2}{3} = \frac{96}{3} = 32$

$x^5 = 32$

Since $2^5 = 32$, we have $x = 2$.

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Now that we have $x=2$, substitute it back into equation (vii) to find $a$:

$\frac{a}{x} = \frac{3}{2}$

$\frac{a}{2} = \frac{3}{2}$

$a = \frac{3}{2} \times 2 = 3$

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We have found the values of $x$, $a$, and $n$. Let's verify these values using equations (i), (ii), and (iii).

Check with (i): $nx^{n-1}a = 5 \times 2^{5-1} \times 3 = 5 \times 2^4 \times 3 = 5 \times 16 \times 3 = 80 \times 3 = 240$. Correct.

Check with (ii): $\frac{n(n-1)}{2}x^{n-2}a^2 = \frac{5(5-1)}{2} \times 2^{5-2} \times 3^2 = \frac{5 \times 4}{2} \times 2^3 \times 9 = 10 \times 8 \times 9 = 80 \times 9 = 720$. Correct.

Check with (iii): $\frac{n(n-1)(n-2)}{6}x^{n-3}a^3 = \frac{5(5-1)(5-2)}{6} \times 2^{5-3} \times 3^3 = \frac{5 \times 4 \times 3}{6} \times 2^2 \times 27 = \frac{60}{6} \times 4 \times 27 = 10 \times 4 \times 27 = 40 \times 27 = 1080$. Correct.

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The values are consistent.

The values are $\mathbf{x=2}$, $\mathbf{a=3}$, and $\mathbf{n=5}$.

Given expansion:

$(1 + a)^n$

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The general term in the expansion of $(1 + x)^n$ is $T_{r+1} = \binom{n}{r} x^r$. In this case, the expansion is $(1+a)^n$, so the general term is $T_{r+1} = \binom{n}{r} a^r$.

The coefficient of the $(r+1)^{\text{th}}$ term is $\binom{n}{r}$.

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Let the three consecutive terms be the $r^{\text{th}}$, $(r+1)^{\text{th}}$, and $(r+2)^{\text{th}}$ terms.

The coefficients of these terms are $\binom{n}{r-1}$, $\binom{n}{r}$, and $\binom{n}{r+1}$ respectively (assuming $r \geq 1$ for the $r^{\text{th}}$ term to exist, and $r+1 \leq n$ for the $(r+2)^{\text{th}}$ term to exist in a non-trivial way). Let's assume the terms start from $T_1$. So the terms are $T_k, T_{k+1}, T_{k+2}$. The coefficients are $\binom{n}{k-1}, \binom{n}{k}, \binom{n}{k+1}$. Let's use $r-1, r, r+1$ for the indices as this is standard.

The coefficients of the three consecutive terms are $\binom{n}{r-1}$, $\binom{n}{r}$, and $\binom{n}{r+1}$.

We are given that the ratio of these coefficients is $1 : 7 : 42$.

So, $\binom{n}{r-1} : \binom{n}{r} : \binom{n}{r+1} = 1 : 7 : 42$.

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From the ratio, we have two equations:

1. $\frac{\binom{n}{r-1}}{\binom{n}{r}} = \frac{1}{7}$

2. $\frac{\binom{n}{r}}{\binom{n}{r+1}} = \frac{7}{42} = \frac{1}{6}$

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We use the property of binomial coefficients: $\frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{n-k+1}{k}$.

Using this property for the first equation:

$\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{n-(r-1)+1}{r-1+1} = \frac{n-r+1+1}{r} = \frac{n-r+2}{r}$

So, $\frac{n-r+2}{r} = 7$ (from $\frac{\binom{n}{r-1}}{\binom{n}{r}} = \frac{1}{7}$) - Error in applying formula. Let's use the basic definition or the inverse property.

The property is $\frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{n-k+1}{k}$.

So, $\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{n-r+1}{r}$.

From the first equation $\frac{\binom{n}{r-1}}{\binom{n}{r}} = \frac{1}{7}$, we have $\frac{\binom{n}{r}}{\binom{n}{r-1}} = 7$.

Therefore, $\frac{n-r+1}{r} = 7$

$n - r + 1 = 7r$

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Using the property for the second equation:

$\frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{n-(r+1)+1}{r+1} = \frac{n-r-1+1}{r+1} = \frac{n-r}{r+1}$

From the second equation $\frac{\binom{n}{r}}{\binom{n}{r+1}} = \frac{1}{6}$, we have $\frac{\binom{n}{r+1}}{\binom{n}{r}} = 6$.

Therefore, $\frac{n-r}{r+1} = 6$

$n - r = 6(r+1)$

$n - r = 6r + 6$

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We have a system of two linear equations with two unknowns, $n$ and $r$:

Equation (i): $n + 1 = 8r$

Equation (ii): $n - 6 = 7r$

Subtract equation (ii) from equation (i):

$(n + 1) - (n - 6) = 8r - 7r$

$n + 1 - n + 6 = r$

$7 = r$

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Now substitute the value of $r=7$ into either equation (i) or (ii) to find $n$. Using equation (i):

$n + 1 = 8r$

$n + 1 = 8(7)$

$n + 1 = 56$

$n = 56 - 1$

$n = 55$

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We should check if the indices $r-1$, $r$, and $r+1$ are valid for $n=55$ and $r=7$. The indices are $7-1=6$, $7$, and $7+1=8$. These are valid indices for an expansion with power 55 (terms from index 0 to 55). The coefficients are $\binom{55}{6}$, $\binom{55}{7}$, and $\binom{55}{8}$.

Let's verify the ratios using $n=55$ and $r=7$ with the property $\frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{n-k+1}{k}$.

Ratio 1: $\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{\binom{55}{7}}{\binom{55}{6}} = \frac{55-7+1}{7} = \frac{49}{7} = 7$. This matches $\frac{7}{1}$.

Ratio 2: $\frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{\binom{55}{8}}{\binom{55}{7}} = \frac{55-8+1}{8} = \frac{48}{8} = 6$. This matches $\frac{42}{7}$.

The values are consistent.

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The value of $n$ is 55.

Given expansion:

$(x + 3)^8$

This is in the form $(a+b)^n$, where $a = x$, $b = 3$, and $n = 8$.

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To Find:

The coefficient of the term $x^5$ in the expansion.

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Solution:

The general term (the $(r+1)^{\text{th}}$ term) in the expansion of $(a+b)^n$ is given by the formula:

$T_{r+1} = \binom{n}{r} a^{n-r} b^r$

Substitute the values $n=8$, $a=x$, and $b=3$:

$T_{r+1} = \binom{8}{r} (x)^{8-r} (3)^r$

$T_{r+1} = \binom{8}{r} 3^r x^{8-r}$

---

We are looking for the term containing $x^5$.

Comparing the exponent of $x$ in the general term with the exponent of $x$ in $x^5$:

$8-r = 5$

$r = 8 - 5 = 3$

So, the term we are interested in is the $(3+1)^{\text{th}}$, which is the 4th term.

---

Substitute $r = 3$ into the general term formula to find the term:

$T_{3+1} = \binom{8}{3} 3^3 x^{8-3}$

$T_4 = \binom{8}{3} 3^3 x^5$

---

Now, calculate the value of the coefficient, which is $\binom{8}{3} 3^3$.

$\binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!} = \frac{8 \times 7 \times 6 \times 5!}{3 \times 2 \times 1 \times 5!} = \frac{8 \times 7 \times 6}{6} = 8 \times 7 = 56$

$3^3 = 3 \times 3 \times 3 = 27$

---

The coefficient of $x^5$ is $\binom{8}{3} 3^3 = 56 \times 27$.

Calculation:

$56 \times 27 = 1512$

---

The term containing $x^5$ is $1512 x^5$.

The coefficient of $x^5$ is $\mathbf{1512}$.

Given expansion:

$(a - 2b)^{12}$

This is in the form $(x+y)^n$, where $x = a$, $y = -2b$, and $n = 12$.

---

To Find:

The coefficient of the term $a^5b^7$ in the expansion.

---

Solution:

The general term (the $(r+1)^{\text{th}}$ term) in the expansion of $(x+y)^n$ is given by the formula:

$T_{r+1} = \binom{n}{r} x^{n-r} y^r$

Substitute the values $n=12$, $x=a$, and $y=-2b$:

$T_{r+1} = \binom{12}{r} (a)^{12-r} (-2b)^r$

$T_{r+1} = \binom{12}{r} a^{12-r} (-2)^r b^r$

$T_{r+1} = \binom{12}{r} (-2)^r a^{12-r} b^r$

---

We are looking for the term containing $a^5b^7$.

Comparing the exponents of $a$ and $b$ in the general term with those in $a^5b^7$:

For the exponent of $a$: $12-r = 5 \implies r = 12 - 5 = 7$.

For the exponent of $b$: $r = 7$.

Both conditions give $r = 7$. So, the term we are interested in is the $(7+1)^{\text{th}}$, which is the 8th term.

---

Substitute $r = 7$ into the general term formula to find the term:

$T_{7+1} = \binom{12}{7} (-2)^7 a^{12-7} b^7$

$T_8 = \binom{12}{7} (-2)^7 a^5 b^7$

---

Now, calculate the value of the coefficient, which is $\binom{12}{7} (-2)^7$.

$\binom{12}{7} = \frac{12!}{7!(12-7)!} = \frac{12!}{7!5!} = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7!}{5 \times 4 \times 3 \times 2 \times 1 \times 7!} = \frac{12 \times 11 \times 10 \times 9 \times 8}{120}$

$\binom{12}{7} = \frac{95040}{120} = 792$

$(-2)^7 = (-1)^7 \times 2^7 = -1 \times 128 = -128$

---

The coefficient of $a^5b^7$ is $\binom{12}{7} (-2)^7 = 792 \times (-128)$.

Calculation:

$792 \times (-128)$

$\begin{array}{ccccccc}& & & 7 & 9 & 2 \\ \times & & & 1 & 2 & 8 \\ \hline && 6 & 3 & 3 & 6 \\ & 1 & 5 & 8 & 4 & \times \\ 7 & 9 & 2 & \times & \times \\ \hline 1 & 0 & 1 & 3 & 7 & 6 \\ \hline \end{array}$

So, $792 \times 128 = 101376$.

Since we are multiplying by -128, the result is negative.

$792 \times (-128) = -101376$.

---

The term containing $a^5b^7$ is $-101376 a^5b^7$.

The coefficient of $a^5b^7$ is $\mathbf{-101376}$.

Given expansion:

$(x^2 - y)^6$

---

To Find:

The general term of the expansion.

---

Solution:

The general term in the expansion of $(a+b)^n$ is the $(r+1)^{\text{th}}$ term, given by the formula:

$T_{r+1} = \binom{n}{r} a^{n-r} b^r$

In the given expansion $(x^2 - y)^6$, we have:

$a = x^2$

$b = -y$

$n = 6$

---

Substitute these values into the general term formula:

$T_{r+1} = \binom{6}{r} (x^2)^{6-r} (-y)^r$

$T_{r+1} = \binom{6}{r} x^{2(6-r)} (-1)^r y^r$

$T_{r+1} = \binom{6}{r} (-1)^r x^{12-2r} y^r$

---

The general term in the expansion of $(x^2 - y)^6$ is $\mathbf{T_{r+1} = \binom{6}{r} (-1)^r x^{12-2r} y^r}$, where $r$ is an integer from 0 to 6 (i.e., $r = 0, 1, 2, 3, 4, 5, 6$).

Given expansion:

$(x^2 - yx)^{12}$

---

To Find:

The general term of the expansion.

---

Solution:

The general term in the expansion of $(a+b)^n$ is the $(r+1)^{\text{th}}$ term, given by the formula:

$T_{r+1} = \binom{n}{r} a^{n-r} b^r$

In the given expansion $(x^2 - yx)^{12}$, we have:

$a = x^2$

$b = -yx$

$n = 12$

---

Substitute these values into the general term formula:

$T_{r+1} = \binom{12}{r} (x^2)^{12-r} (-yx)^r$

$T_{r+1} = \binom{12}{r} x^{2(12-r)} (-1)^r y^r x^r$

$T_{r+1} = \binom{12}{r} (-1)^r x^{24-2r} y^r x^r$

Combine the terms with $x$ by adding their exponents:

$T_{r+1} = \binom{12}{r} (-1)^r x^{24-2r+r} y^r$

$T_{r+1} = \binom{12}{r} (-1)^r x^{24-r} y^r$

---

The general term in the expansion of $(x^2 - yx)^{12}$ is $\mathbf{T_{r+1} = \binom{12}{r} (-1)^r x^{24-r} y^r}$, where $r$ is an integer from 0 to 12 (i.e., $r = 0, 1, 2, ..., 12$).

Given expansion:

$(x - 2y)^{12}$

---

To Find:

The 4^th term in the expansion.

---

Solution:

The general term (the $(r+1)^{\text{th}}$ term) in the expansion of $(a+b)^n$ is given by the formula:

$T_{r+1} = \binom{n}{r} a^{n-r} b^r$

In the given expansion $(x - 2y)^{12}$, we have:

$a = x$

$b = -2y$

$n = 12$

---

We want to find the 4^th term, which means $r+1 = 4$.

So, $r = 4 - 1 = 3$.

---

Substitute $r=3$, $n=12$, $a=x$, and $b=-2y$ into the general term formula:

$T_{3+1} = \binom{12}{3} (x)^{12-3} (-2y)^3$

$T_4 = \binom{12}{3} x^9 (-2)^3 y^3$

---

Now, calculate the values of the binomial coefficient and the power of -2:

$\binom{12}{3} = \frac{12!}{3!(12-3)!} = \frac{12!}{3!9!} = \frac{12 \times 11 \times 10 \times 9!}{3 \times 2 \times 1 \times 9!} = \frac{12 \times 11 \times 10}{6} = 2 \times 11 \times 10 = 220$

$(-2)^3 = (-1)^3 \times 2^3 = -1 \times 8 = -8$

---

Substitute these values back into the expression for $T_4$:

$T_4 = 220 \times x^9 \times (-8) \times y^3$

$T_4 = 220 \times (-8) \times x^9 y^3$

$T_4 = -1760 x^9 y^3$

---

The 4^th term in the expansion of $(x - 2y)^{12}$ is $\mathbf{-1760 x^9 y^3}$.

Given expansion:

$\left( 9x-\frac{1}{3\sqrt{x}} \right)^{18}$

---

To Find:

The 13^th term in the expansion.

---

Solution:

The general term (the $(r+1)^{\text{th}}$ term) in the expansion of $(a+b)^n$ is given by the formula:

$T_{r+1} = \binom{n}{r} a^{n-r} b^r$

In the given expansion $\left( 9x-\frac{1}{3\sqrt{x}} \right)^{18}$, we have:

$a = 9x$

$b = -\frac{1}{3\sqrt{x}}$

$n = 18$

---

We want to find the 13^th term, which means $r+1 = 13$.

So, $r = 13 - 1 = 12$.

---

Substitute $r=12$, $n=18$, $a=9x$, and $b=-\frac{1}{3\sqrt{x}}$ into the general term formula:

$T_{12+1} = \binom{18}{12} (9x)^{18-12} \left(-\frac{1}{3\sqrt{x}}\right)^{12}$

$T_{13} = \binom{18}{12} (9x)^6 \left(-\frac{1}{3x^{1/2}}\right)^{12}$

$T_{13} = \binom{18}{12} 9^6 x^6 \left(\frac{(-1)^{12}}{3^{12} (x^{1/2})^{12}}\right)$

$T_{13} = \binom{18}{12} 9^6 x^6 \left(\frac{1}{3^{12} x^{12/2}}\right)$

$T_{13} = \binom{18}{12} 9^6 x^6 \left(\frac{1}{3^{12} x^6}\right)$

---

Simplify the powers of 9 and 3. Since $9 = 3^2$, $9^6 = (3^2)^6 = 3^{12}$.

$T_{13} = \binom{18}{12} 3^{12} x^6 \left(\frac{1}{3^{12} x^6}\right)$

$T_{13} = \binom{18}{12} \frac{3^{12}}{3^{12}} \frac{x^6}{x^6}$

$T_{13} = \binom{18}{12} \times 1 \times 1$

$T_{13} = \binom{18}{12}$

---

Now, calculate the value of the binomial coefficient $\binom{18}{12}$.

We know that $\binom{n}{k} = \binom{n}{n-k}$. So, $\binom{18}{12} = \binom{18}{18-12} = \binom{18}{6}$.

$\binom{18}{6} = \frac{18!}{6!(18-6)!} = \frac{18!}{6!12!} = \frac{18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12!}{6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 12!}$

$\binom{18}{6} = \frac{18 \times 17 \times 16 \times 15 \times 14 \times 13}{6 \times 5 \times 4 \times 3 \times 2 \times 1}$

The denominator is $6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$.

Numerator: $18 \times 17 \times 16 \times 15 \times 14 \times 13 = 13366080$.

$\binom{18}{6} = \frac{13366080}{720} = 18564$

---

The 13^th term is a constant term (independent of $x$) and its value is 18564.

---

The 13^th term in the expansion of $\left( 9x-\frac{1}{3\sqrt{x}} \right)^{18}$ is $\mathbf{18564}$.

Given expansion:

$\left( 3 - \frac{x^3}{6} \right)^7$

---

To Find:

The middle terms in the expansion.

---

Solution:

The expansion is of the form $(a+b)^n$, where $a = 3$, $b = -\frac{x^3}{6}$, and $n = 7$.

The total number of terms in the expansion is $n+1 = 7+1 = 8$.

---

Since the number of terms (8) is even, there are two middle terms.

The positions of the middle terms are $\left(\frac{n}{2}+1\right)^{\text{th}}$ term and $\left(\frac{n}{2}+2\right)^{\text{th}}$ term.

Here $n=7$, which is odd. The positions are $\left(\frac{n+1}{2}\right)^{\text{th}}$ term and $\left(\frac{n+1}{2}+1\right)^{\text{th}}$ term.

Middle term positions = $\left(\frac{7+1}{2}\right)^{\text{th}}$ term and $\left(\frac{7+1}{2}+1\right)^{\text{th}}$ term.

Middle term positions = $(4)^{\text{th}}$ term and $(4+1)^{\text{th}}$ term = 4^th term and 5^th term.

---

The general term (the $(r+1)^{\text{th}}$ term) in the expansion of $(a+b)^n$ is given by the formula:

$T_{r+1} = \binom{n}{r} a^{n-r} b^r$

Substitute the values $n=7$, $a=3$, and $b=-\frac{x^3}{6}$:

$T_{r+1} = \binom{7}{r} (3)^{7-r} \left(-\frac{x^3}{6}\right)^r$

$T_{r+1} = \binom{7}{r} 3^{7-r} (-1)^r \frac{(x^3)^r}{6^r}$

$T_{r+1} = \binom{7}{r} (-1)^r 3^{7-r} \frac{x^{3r}}{6^r}$

Since $6 = 2 \times 3$, $6^r = 2^r \times 3^r$.

$T_{r+1} = \binom{7}{r} (-1)^r 3^{7-r} \frac{x^{3r}}{2^r 3^r}$

$T_{r+1} = \binom{7}{r} (-1)^r \frac{3^{7-r}}{3^r} \frac{x^{3r}}{2^r}$

$T_{r+1} = \binom{7}{r} (-1)^r 3^{7-r-r} \frac{x^{3r}}{2^r}$

$T_{r+1} = \binom{7}{r} (-1)^r \frac{3^{7-2r}}{2^r} x^{3r}$

---

First middle term is the 4^th term ($r=3$):

$T_4 = T_{3+1} = \binom{7}{3} (-1)^3 \frac{3^{7-2(3)}}{2^3} x^{3(3)}$

$T_4 = \binom{7}{3} (-1) \frac{3^{7-6}}{8} x^9$

$T_4 = \binom{7}{3} (-1) \frac{3^1}{8} x^9$

Calculate $\binom{7}{3}$: $\binom{7}{3} = \frac{7!}{3!4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.

$T_4 = 35 \times (-1) \times \frac{3}{8} \times x^9$

$T_4 = -35 \times \frac{3}{8} x^9 = -\frac{105}{8} x^9$

---

Second middle term is the 5^th term ($r=4$):

$T_5 = T_{4+1} = \binom{7}{4} (-1)^4 \frac{3^{7-2(4)}}{2^4} x^{3(4)}$

$T_5 = \binom{7}{4} (1) \frac{3^{7-8}}{16} x^{12}$

$T_5 = \binom{7}{4} \frac{3^{-1}}{16} x^{12}$

Calculate $\binom{7}{4}$: $\binom{7}{4} = \binom{7}{7-4} = \binom{7}{3} = 35$.

$T_5 = 35 \times \frac{1}{3 \times 16} x^{12}$

$T_5 = 35 \times \frac{1}{48} x^{12} = \frac{35}{48} x^{12}$

---

The middle terms are the 4^th and 5^th terms.

The 4^th term is $\mathbf{-\frac{105}{8} x^9}$.

The 5^th term is $\mathbf{\frac{35}{48} x^{12}}$.

Given expansion:

$\left( \frac{x}{3}+9y \right)^{10}$

---

To Find:

The middle term in the expansion.

---

Solution:

The expansion is of the form $(a+b)^n$, where $a = \frac{x}{3}$, $b = 9y$, and $n = 10$.

The total number of terms in the expansion is $n+1 = 10+1 = 11$.

---

Since the number of terms (11) is odd, there is exactly one middle term.

The position of the middle term is $\left(\frac{n}{2}+1\right)^{\text{th}}$ term. No, this is for even $n+1$.

The position of the middle term is $\left(\frac{\text{Total number of terms} + 1}{2}\right)^{\text{th}}$ term.

Middle term position = $\left(\frac{11+1}{2}\right)^{\text{th}}$ term = $\left(\frac{12}{2}\right)^{\text{th}}$ term = 6^th term.

---

The general term (the $(r+1)^{\text{th}}$ term) in the expansion of $(a+b)^n$ is given by the formula:

$T_{r+1} = \binom{n}{r} a^{n-r} b^r$

Substitute the values $n=10$, $a=\frac{x}{3}$, and $b=9y$:

$T_{r+1} = \binom{10}{r} \left(\frac{x}{3}\right)^{10-r} (9y)^r$

$T_{r+1} = \binom{10}{r} \frac{x^{10-r}}{3^{10-r}} 9^r y^r$

Since $9 = 3^2$, $9^r = (3^2)^r = 3^{2r}$.

$T_{r+1} = \binom{10}{r} \frac{x^{10-r}}{3^{10-r}} 3^{2r} y^r$

$T_{r+1} = \binom{10}{r} 3^{2r - (10-r)} x^{10-r} y^r$

$T_{r+1} = \binom{10}{r} 3^{2r - 10 + r} x^{10-r} y^r$

$T_{r+1} = \binom{10}{r} 3^{3r - 10} x^{10-r} y^r$

---

The middle term is the 6^th term, which means $r+1 = 6$.

So, $r = 6 - 1 = 5$.

---

Substitute $r=5$ into the general term formula:

$T_6 = T_{5+1} = \binom{10}{5} 3^{3(5) - 10} x^{10-5} y^5$

$T_6 = \binom{10}{5} 3^{15 - 10} x^5 y^5$

$T_6 = \binom{10}{5} 3^5 x^5 y^5$

---

Now, calculate the values of the binomial coefficient and the power of 3:

$\binom{10}{5} = \frac{10!}{5!(10-5)!} = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{5 \times 4 \times 3 \times 2 \times 1 \times 5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{120}$

$\binom{10}{5} = \frac{30240}{120} = 252$

$3^5 = 3 \times 3 \times 3 \times 3 \times 3 = 9 \times 9 \times 3 = 81 \times 3 = 243$

---

Substitute these values back into the expression for $T_6$:

$T_6 = 252 \times 243 \times x^5 y^5$

Calculate $252 \times 243$:

$\begin{array}{cccccc}& & & 2 & 5 & 2 \\ \times & & & 2 & 4 & 3 \\ \hline && 7 & 5 & 6 \\ & 1 & 0 & 0 & 8 & \times \\ 5 & 0 & 4 & \times & \times \\ \hline 6 & 1 & 2 & 4 & 3 & 6 \\ \hline \end{array}$

$252 \times 243 = 61236$.

---

The middle term is $61236 x^5 y^5$.

The middle term in the expansion of $\left( \frac{x}{3}+9y \right)^{10}$ is $\mathbf{61236 x^5 y^5}$.

Given expansion:

$(1 + a)^{m+n}$

This is in the form $(x+y)^N$, where $x=1$, $y=a$, and $N = m+n$.

---

To Prove:

The coefficient of $a^m$ is equal to the coefficient of $a^n$ in the expansion of $(1+a)^{m+n}$.

---

Proof:

The general term in the expansion of $(1+y)^N$ is $T_{r+1} = \binom{N}{r} (1)^{N-r} y^r = \binom{N}{r} y^r$.

In our case, $y=a$ and $N=m+n$. So, the general term is:

$T_{r+1} = \binom{m+n}{r} a^r$

---

We want to find the coefficient of $a^m$. This occurs when the exponent of $a$ in the general term is $m$.

$r = m$

The coefficient of $a^m$ is obtained by setting $r=m$ in the general term's coefficient $\binom{m+n}{r}$.

Coefficient of $a^m = \binom{m+n}{m}$.

---

We want to find the coefficient of $a^n$. This occurs when the exponent of $a$ in the general term is $n$.

$r = n$

The coefficient of $a^n$ is obtained by setting $r=n$ in the general term's coefficient $\binom{m+n}{r}$.

Coefficient of $a^n = \binom{m+n}{n}$.

---

We need to show that $\binom{m+n}{m} = \binom{m+n}{n}$.

Recall the property of binomial coefficients: $\binom{N}{k} = \binom{N}{N-k}$.

In our case, $N = m+n$ and we want to show $\binom{m+n}{m} = \binom{m+n}{n}$.

Let $k = m$. Then $N-k = (m+n) - m = n$.

Using the property, $\binom{m+n}{m} = \binom{m+n}{(m+n)-m} = \binom{m+n}{n}$.

---

Therefore, the coefficient of $a^m$, which is $\binom{m+n}{m}$, is equal to the coefficient of $a^n$, which is $\binom{m+n}{n}$.

Thus, the coefficients of $a^m$ and $a^n$ in the expansion of $(1+a)^{m+n}$ are equal.

Hence Proved.

Given expansion:

$(x + 1)^n$

---

The general term in the expansion of $(a+b)^n$ is $T_{k+1} = \binom{n}{k} a^{n-k} b^k$.

In this case, $a=x$, $b=1$. The general term is $T_{k+1} = \binom{n}{k} x^{n-k} 1^k = \binom{n}{k} x^{n-k}$.

The coefficient of the $(k+1)^{\text{th}}$ term is $\binom{n}{k}$.

---

We are given the coefficients of the $(r-1)^{\text{th}}$, $r^{\text{th}}$, and $(r+1)^{\text{th}}$ terms.

The $(r-1)^{\text{th}}$ term is $T_{r-1}$. Its coefficient is $\binom{n}{(r-1)-1} = \binom{n}{r-2}$. (Assuming $r-1 \geq 1$, so $r \geq 2$)

The $r^{\text{th}}$ term is $T_r$. Its coefficient is $\binom{n}{r-1}$. (Assuming $r \geq 1$)

The $(r+1)^{\text{th}}$ term is $T_{r+1}$. Its coefficient is $\binom{n}{r}$. (Assuming $r \geq 0$)

For all three terms to exist with non-zero coefficients, we need $r-2 \geq 0$, so $r \geq 2$, and $r \leq n$, and $r+1 \leq n+1$. Thus, $2 \leq r \leq n$.

The coefficients are $\binom{n}{r-2}$, $\binom{n}{r-1}$, and $\binom{n}{r}$.

---

We are given that these coefficients are in the ratio $1 : 3 : 5$.

So, $\binom{n}{r-2} : \binom{n}{r-1} : \binom{n}{r} = 1 : 3 : 5$.

---

From the ratio, we have two equations:

1. $\frac{\binom{n}{r-2}}{\binom{n}{r-1}} = \frac{1}{3}$

2. $\frac{\binom{n}{r-1}}{\binom{n}{r}} = \frac{3}{5}$

---

We use the property of binomial coefficients: $\frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{n-k+1}{k}$.

Using this property for the first equation (with $k=r-1$):

$\frac{\binom{n}{r-1}}{\binom{n}{r-2}} = \frac{n-(r-1)+1}{r-1} = \frac{n-r+1+1}{r-1} = \frac{n-r+2}{r-1}$

From $\frac{\binom{n}{r-2}}{\binom{n}{r-1}} = \frac{1}{3}$, we have $\frac{\binom{n}{r-1}}{\binom{n}{r-2}} = 3$.

Therefore, $\frac{n-r+2}{r-1} = 3$

$n - r + 2 = 3(r-1)$

$n - r + 2 = 3r - 3$

$n + 2 + 3 = 3r + r$

---

Using the property for the second equation (with $k=r$):

$\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{n-r+1}{r}$

From $\frac{\binom{n}{r-1}}{\binom{n}{r}} = \frac{3}{5}$, we have $\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{5}{3}$.

Therefore, $\frac{n-r+1}{r} = \frac{5}{3}$

$3(n - r + 1) = 5r$

$3n - 3r + 3 = 5r$

$3n + 3 = 5r + 3r$

---

We now have a system of two linear equations with two unknowns, $n$ and $r$:

Equation (i): $n + 5 = 4r$

Equation (ii): $3n + 3 = 8r$

Multiply equation (i) by 2 to make the coefficient of $r$ the same as in equation (ii):

$2(n + 5) = 2(4r)$

$2n + 10 = 8r$

Now equate the expressions for $8r$ from equation (ii) and equation (iii):

$3n + 3 = 2n + 10$

$3n - 2n = 10 - 3$

---

Now substitute the value of $n=7$ into equation (i) to find $r$:

$n + 5 = 4r$

$7 + 5 = 4r$

$12 = 4r$

$r = \frac{12}{4} = 3$

---

We need to check if the value of $r=3$ satisfies the condition $r \geq 2$. Yes, $3 \geq 2$. We also need to check if $r \leq n$. Yes, $3 \leq 7$. The coefficients are $\binom{7}{3-2}=\binom{7}{1}$, $\binom{7}{3-1}=\binom{7}{2}$, and $\binom{7}{3}=\binom{7}{3}$.

Let's calculate these coefficients and check their ratio:

$\binom{7}{1} = 7$

$\binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21$

$\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$

The ratio of the coefficients $\binom{7}{1} : \binom{7}{2} : \binom{7}{3}$ is $7 : 21 : 35$.

Divide by the greatest common divisor, which is 7:

$7/7 : 21/7 : 35/7 = 1 : 3 : 5$.

This matches the given ratio.

---

The values are consistent.

The values are $\mathbf{n=7}$ and $\mathbf{r=3}$.

To Prove:

The coefficient of $x^n$ in the expansion of $(1 + x)^{2n}$ is twice the coefficient of $x^n$ in the expansion of $(1 + x)^{2n – 1}$.

Mathematically, we need to show that $\text{Coefficient}(x^n \text{ in } (1+x)^{2n}) = 2 \times \text{Coefficient}(x^n \text{ in } (1+x)^{2n-1})$.

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Proof:

The general term in the binomial expansion of $(1+x)^N$ is given by $T_{k+1} = \binom{N}{k} 1^{N-k} x^k = \binom{N}{k} x^k$. The coefficient of $x^k$ is $\binom{N}{k}$.

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Consider the expansion of $(1+x)^{2n}$. Here, the power of the expansion is $N = 2n$.

We are looking for the coefficient of $x^n$, which means the exponent of $x$ is $k=n$.

The coefficient of $x^n$ in $(1+x)^{2n}$ is $\binom{2n}{n}$.

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Consider the expansion of $(1+x)^{2n-1}$. Here, the power of the expansion is $N = 2n-1$.

We are looking for the coefficient of $x^n$, which means the exponent of $x$ is $k=n$.

The coefficient of $x^n$ in $(1+x)^{2n-1}$ is $\binom{2n-1}{n}$.

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We need to prove that $\binom{2n}{n} = 2 \times \binom{2n-1}{n}$.

Let's start with the left side of the equation and use the definition of the binomial coefficient $\binom{N}{k} = \frac{N!}{k!(N-k)!}$:

Left Hand Side (LHS) = $\binom{2n}{n} = \frac{(2n)!}{n!(2n-n)!} = \frac{(2n)!}{n!n!}$

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Now consider the right side of the equation:

Right Hand Side (RHS) = $2 \times \binom{2n-1}{n}$

Using the definition of the binomial coefficient for $\binom{2n-1}{n}$:

$\binom{2n-1}{n} = \frac{(2n-1)!}{n!((2n-1)-n)!} = \frac{(2n-1)!}{n!(n-1)!}$

So, RHS = $2 \times \frac{(2n-1)!}{n!(n-1)!}$

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We want to show that LHS = RHS, i.e., $\frac{(2n)!}{n!n!} = 2 \times \frac{(2n-1)!}{n!(n-1)!}$.

Let's manipulate the expression for LHS:

LHS = $\frac{(2n)!}{n!n!}$

We know that $(2n)! = (2n) \times (2n-1)!$.

LHS = $\frac{2n \times (2n-1)!}{n!n!}$

We also know that $n! = n \times (n-1)!$. Let's substitute this for one of the $n!$ terms in the denominator:

LHS = $\frac{2n \times (2n-1)!}{n! \times (n \times (n-1)!)}$

Rearrange the terms in the denominator and numerator:

LHS = $\frac{2n}{n} \times \frac{(2n-1)!}{n! \times (n-1)!}$

Simplify $\frac{2n}{n} = 2$ (assuming $n$ is a positive integer, so $n \neq 0$):

LHS = $2 \times \frac{(2n-1)!}{n!(n-1)!}$

We recognize the term $\frac{(2n-1)!}{n!(n-1)!}$ as $\binom{2n-1}{n}$.

LHS = $2 \times \binom{2n-1}{n}$

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Thus, we have shown that LHS = $2 \times \binom{2n-1}{n}$, which is equal to the RHS.

$\binom{2n}{n} = 2 \times \binom{2n-1}{n}$

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This proves that the coefficient of $x^n$ in the expansion of $(1+x)^{2n}$ is twice the coefficient of $x^n$ in the expansion of $(1+x)^{2n-1}$.

Hence Proved.

Given expansion:

$(1 + x)^m$

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To Find:

A positive value of $m$ such that the coefficient of $x^2$ in the expansion is 6.

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Solution:

The general term in the binomial expansion of $(1+x)^m$ is given by $T_{r+1} = \binom{m}{r} 1^{m-r} x^r = \binom{m}{r} x^r$.

The coefficient of $x^r$ is $\binom{m}{r}$.

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We are looking for the coefficient of $x^2$. This means $r=2$.

The coefficient of $x^2$ in the expansion of $(1+x)^m$ is $\binom{m}{2}$.

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We are given that this coefficient is equal to 6.

So, we have the equation:

$\binom{m}{2} = 6$

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Using the definition of the binomial coefficient $\binom{m}{k} = \frac{m(m-1)...(m-k+1)}{k!}$ for a real number $m$ and non-negative integer $k$. For integer $m$, $\binom{m}{k} = \frac{m!}{k!(m-k)!}$.

For $\binom{m}{2}$, where $m$ is a positive value (and likely an integer for the binomial coefficient to be easily defined and non-zero for $k=2$), we have:

$\binom{m}{2} = \frac{m(m-1)}{2!}$

$\binom{m}{2} = \frac{m(m-1)}{2 \times 1} = \frac{m(m-1)}{2}$

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Substitute this into the equation $\binom{m}{2} = 6$:

$\frac{m(m-1)}{2} = 6$

$m(m-1) = 6 \times 2$

$m(m-1) = 12$

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Expand the left side and form a quadratic equation:

$m^2 - m = 12$

$m^2 - m - 12 = 0$

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We can solve this quadratic equation by factoring or using the quadratic formula. Let's try factoring.

We need two numbers that multiply to -12 and add to -1. These numbers are -4 and 3.

$(m - 4)(m + 3) = 0$

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Setting each factor to zero gives the possible values for $m$:

$m - 4 = 0 \implies m = 4$

$m + 3 = 0 \implies m = -3$

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The question asks for a positive value of $m$.

From the two solutions, $m=4$ is positive, and $m=-3$ is negative.

Therefore, the positive value of $m$ is 4.

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Let's verify for $m=4$. The expansion is $(1+x)^4$.

The coefficient of $x^2$ in $(1+x)^4$ is $\binom{4}{2}$.

$\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3 \times 2!}{2 \times 1 \times 2!} = \frac{4 \times 3}{2} = 6$.

This matches the given condition.

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The positive value of m is $\mathbf{4}$.

Given expansion:

$\left( \frac{3}{2}x^2 - \frac{1}{3x} \right)^6$

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To Find:

The term independent of $x$ in the expansion.

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Solution:

The expansion is of the form $(a+b)^n$, where $a = \frac{3}{2}x^2$, $b = -\frac{1}{3x}$, and $n = 6$.

The general term (the $(r+1)^{\text{th}}$ term) in the expansion of $(a+b)^n$ is given by the formula:

$T_{r+1} = \binom{n}{r} a^{n-r} b^r$

Substitute the values $n=6$, $a=\frac{3}{2}x^2$, and $b=-\frac{1}{3x}$:

$T_{r+1} = \binom{6}{r} \left(\frac{3}{2}x^2\right)^{6-r} \left(-\frac{1}{3x}\right)^r$

$T_{r+1} = \binom{6}{r} \left(\frac{3}{2}\right)^{6-r} (x^2)^{6-r} \left(-\frac{1}{3}\right)^r \left(\frac{1}{x}\right)^r$

$T_{r+1} = \binom{6}{r} \left(\frac{3}{2}\right)^{6-r} x^{2(6-r)} \left(-\frac{1}{3}\right)^r x^{-r}$

$T_{r+1} = \binom{6}{r} \left(\frac{3}{2}\right)^{6-r} \left(-\frac{1}{3}\right)^r x^{12-2r} x^{-r}$

Combine the terms with $x$ by adding their exponents:

$T_{r+1} = \binom{6}{r} \left(\frac{3}{2}\right)^{6-r} \left(-\frac{1}{3}\right)^r x^{12-2r-r}$

$T_{r+1} = \binom{6}{r} \left(\frac{3}{2}\right)^{6-r} \left(-\frac{1}{3}\right)^r x^{12-3r}$

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The term independent of $x$ is the term where the exponent of $x$ is 0.

Set the exponent of $x$ in the general term equal to 0:

$12 - 3r = 0$

$3r = 12$

$r = \frac{12}{3} = 4$

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So, the term independent of $x$ is the $(4+1)^{\text{th}}$, which is the 5th term.

Substitute $r=4$ into the expression for the general term, but only the coefficient part (the part without $x^{12-3r}$):

The term independent of $x$ is $T_5 = \binom{6}{4} \left(\frac{3}{2}\right)^{6-4} \left(-\frac{1}{3}\right)^4$.

$T_5 = \binom{6}{4} \left(\frac{3}{2}\right)^2 \left(-\frac{1}{3}\right)^4$

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Now, calculate the values:

$\binom{6}{4} = \binom{6}{6-4} = \binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15$

$\left(\frac{3}{2}\right)^2 = \frac{3^2}{2^2} = \frac{9}{4}$

$\left(-\frac{1}{3}\right)^4 = (-1)^4 \times \left(\frac{1}{3}\right)^4 = 1 \times \frac{1^4}{3^4} = \frac{1}{81}$

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Substitute these values into the expression for $T_5$:

$T_5 = 15 \times \frac{9}{4} \times \frac{1}{81}$

Cancel common factors:

$T_5 = \frac{15}{1} \times \frac{9}{4} \times \frac{1}{81} = \frac{15 \times \cancel{9}^{1}}{4 \times \cancel{81}^{9}} = \frac{\cancel{15}^{5} \times 1}{4 \times \cancel{9}^{3}} = \frac{5 \times 1}{4 \times 3} = \frac{5}{12}$

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The term independent of $x$ is $\mathbf{\frac{5}{12}}$.

Given expansion:

$(1 + a)^n$

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The general term in the expansion of $(1 + x)^n$ is $T_{k+1} = \binom{n}{k} x^k$. In this case, the expansion is $(1+a)^n$, so the general term is $T_{k+1} = \binom{n}{k} a^k$.

The coefficient of $a^k$ is $\binom{n}{k}$.

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The coefficients of $a^{r-1}$, $a^r$, and $a^{r+1}$ are:

Coefficient of $a^{r-1}$ is $\binom{n}{r-1}$. (Assuming $r-1 \geq 0$, so $r \geq 1$)

Coefficient of $a^{r}$ is $\binom{n}{r}$. (Assuming $r \geq 0$)

Coefficient of $a^{r+1}$ is $\binom{n}{r+1}$. (Assuming $r+1 \leq n$, so $r \leq n-1$)

For all three coefficients to be non-zero and well-defined in the standard binomial expansion, we need $1 \leq r \leq n-1$. This implies $n \geq 2$.

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We are given that these coefficients are in arithmetic progression (AP).

If three numbers $A, B, C$ are in AP, then the middle term is the average of the other two, i.e., $B = \frac{A+C}{2}$, which means $2B = A+C$.

So, $2 \times \binom{n}{r} = \binom{n}{r-1} + \binom{n}{r+1}$.

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Express the binomial coefficients using the definition $\binom{n}{k} = \frac{n!}{k!(n-k)!}$:

$\binom{n}{r-1} = \frac{n!}{(r-1)!(n-(r-1))!} = \frac{n!}{(r-1)!(n-r+1)!}$

$\binom{n}{r} = \frac{n!}{r!(n-r)!}$

$\binom{n}{r+1} = \frac{n!}{(r+1)!(n-(r+1))!} = \frac{n!}{(r+1)!(n-r-1)!}$

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Substitute these into equation (i):

$2 \times \frac{n!}{r!(n-r)!} = \frac{n!}{(r-1)!(n-r+1)!} + \frac{n!}{(r+1)!(n-r-1)!}$

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Divide both sides by $n!$ (assuming $n! \neq 0$, which is true for integer $n \geq 0$):

$\frac{2}{r!(n-r)!} = \frac{1}{(r-1)!(n-r+1)!} + \frac{1}{(r+1)!(n-r-1)!}$

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Express the factorials in terms of the smallest factorials in the denominator:

$r! = r \times (r-1)!$

$(r+1)! = (r+1) \times r \times (r-1)!$

$(n-r)! = (n-r) \times (n-r-1)!$

$(n-r+1)! = (n-r+1) \times (n-r) \times (n-r-1)!$

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Substitute these expansions into the equation:

$\frac{2}{r(r-1)!(n-r)(n-r-1)!} = \frac{1}{(r-1)!(n-r+1)(n-r)(n-r-1)!} + \frac{1}{(r+1)r(r-1)!(n-r-1)!}$

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Multiply both sides by $(r-1)!(n-r-1)!$ (assuming they are non-zero):

$\frac{2}{r(n-r)} = \frac{1}{(n-r+1)(n-r)} + \frac{1}{(r+1)r}$

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Find a common denominator on the right side, which is $(n-r+1)(n-r)(r+1)r$:

$\frac{2}{r(n-r)} = \frac{(r+1)r + (n-r+1)(n-r)}{(n-r+1)(n-r)(r+1)r}$

$\frac{2}{r(n-r)} = \frac{r^2+r + n^2 - nr + n - nr + r^2 - r}{(n-r+1)(n-r)(r+1)r}$

$\frac{2}{r(n-r)} = \frac{r^2+r + n^2 - 2nr + n + r^2 - r}{(n-r+1)(n-r)(r+1)r}$

$\frac{2}{r(n-r)} = \frac{2r^2 + n^2 - 2nr + n}{(n-r+1)(n-r)(r+1)r}$

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Cross-multiply:

$2(n-r+1)(n-r)(r+1)r = r(n-r)(2r^2 + n^2 - 2nr + n)$

Assuming $r \neq 0$ and $n-r \neq 0$ (which is true from $1 \leq r \leq n-1$), we can divide both sides by $r(n-r)$:

$2(n-r+1)(r+1) = 2r^2 + n^2 - 2nr + n$

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Expand the left side:

$2(nr + n + (-r)r + (-r) + 1r + 1)$

$2(nr + n - r^2 - r + r + 1)$

$2(nr + n - r^2 + 1)$

$2nr + 2n - 2r^2 + 2$

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So, $2nr + 2n - 2r^2 + 2 = 2r^2 + n^2 - 2nr + n$

Move all terms to one side (e.g., to the right side):

$0 = (n^2 - 2nr + n + 2r^2) - (2nr + 2n - 2r^2 + 2)$

$0 = n^2 - 2nr + n + 2r^2 - 2nr - 2n + 2r^2 - 2$

Combine like terms:

$0 = n^2 + (-2nr - 2nr) + (n - 2n) + (2r^2 + 2r^2) - 2$

$0 = n^2 - 4nr - n + 4r^2 - 2$

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Rearrange the terms to match the desired form:

$n^2 - n - 4nr + 4r^2 - 2 = 0$

$n^2 - n(1 + 4r) + 4r^2 - 2 = 0$

$n^2 - n(4r + 1) + 4r^2 - 2 = 0$

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This matches the equation we needed to prove.

Hence Proved.

Given expansions:

$(1 + x)^{2n}$ and $(1 + x)^{2n-1}$

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To Show:

The coefficient of the middle term in $(1 + x)^{2n}$ is equal to the sum of the coefficients of the two middle terms in $(1 + x)^{2n – 1}$.

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Proof:

Consider the expansion of $(1+x)^N$. The general term is $T_{k+1} = \binom{N}{k} 1^{N-k} x^k = \binom{N}{k} x^k$. The coefficient of $x^k$ is $\binom{N}{k}$.

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Expansion of $(1 + x)^{2n}$:

The power is $N_1 = 2n$. The number of terms is $2n+1$. Since $2n+1$ is odd, there is one middle term.

The position of the middle term is $\left(\frac{(2n+1)+1}{2}\right)^{\text{th}} = (n+1)^{\text{th}}$.

The middle term is $T_{n+1}$. The coefficient of the middle term is obtained by setting $k=n$ and $N=2n$ in $\binom{N}{k}$.

Coefficient of the middle term in $(1+x)^{2n} = \binom{2n}{n}$.

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Expansion of $(1 + x)^{2n-1}$:

The power is $N_2 = 2n-1$. The number of terms is $(2n-1)+1 = 2n$. Since $2n$ is even, there are two middle terms.

The positions of the two middle terms are $\left(\frac{2n}{2}\right)^{\text{th}} = n^{\text{th}}$ and $\left(\frac{2n}{2}+1\right)^{\text{th}} = (n+1)^{\text{th}}$.

The coefficients of these terms are obtained by setting $N=2n-1$ and $k=n-1$ (for the $n^{\text{th}}$ term) and $k=n$ (for the $(n+1)^{\text{th}}$ term) in $\binom{N}{k}$.

Coefficient of the first middle term ($n^{\text{th}}$ term) in $(1+x)^{2n-1} = \binom{2n-1}{n-1}$.

Coefficient of the second middle term ($(n+1)^{\text{th}}$ term) in $(1+x)^{2n-1} = \binom{2n-1}{n}$.

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We need to show that Coefficient$_1$ = Coefficient$_2$.

That is, we need to prove $\binom{2n}{n} = \binom{2n-1}{n-1} + \binom{2n-1}{n}$.

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This is a well-known identity in combinatorics and Pascal's triangle: $\binom{N}{k} = \binom{N-1}{k-1} + \binom{N-1}{k}$.

Let $N = 2n$ and $k = n$. Substituting these values into the identity:

$\binom{2n}{n} = \binom{2n-1}{n-1} + \binom{2n-1}{n}$.

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Alternatively, we can prove this identity using the definition of binomial coefficients $\binom{N}{k} = \frac{N!}{k!(N-k)!}$:

Consider the right side of the identity:

RHS = $\binom{2n-1}{n-1} + \binom{2n-1}{n}$

RHS = $\frac{(2n-1)!}{(n-1)!((2n-1)-(n-1))!} + \frac{(2n-1)!}{n!((2n-1)-n)!}$

RHS = $\frac{(2n-1)!}{(n-1)!(2n-n) \times (n-1)!} + \frac{(2n-1)!}{n!(n-1)!}$ No, factorial expansion error.

RHS = $\frac{(2n-1)!}{(n-1)!(2n-1-n+1)!} + \frac{(2n-1)!}{n!(2n-1-n)!}$

RHS = $\frac{(2n-1)!}{(n-1)!n!} + \frac{(2n-1)!}{n!(n-1)!}$

To add these fractions, we use the common denominator $n!(n-1)!$. However, the terms are already over a common denominator (since $n!(n-1)! = (n-1)!n!$). Let's find a common denominator that includes $n!$ for both terms.

We know $n! = n \times (n-1)!$. So, $(n-1)! = \frac{n!}{n}$.

RHS = $\frac{(2n-1)!}{\left(\frac{n!}{n}\right)n!} + \frac{(2n-1)!}{n!(n-1)!}$ No, this is not the correct way to find a common denominator.

Let the common denominator be $n!(n-1)!$.

RHS = $\frac{(2n-1)!}{(n-1)!n!} + \frac{(2n-1)!}{n!(n-1)!}$

We can factor out $(2n-1)!$ and $\frac{1}{(n-1)!}$:

RHS = $(2n-1)! \frac{1}{(n-1)!} \left( \frac{1}{n!} + \frac{1}{n!} \right)$ No, the second denominator is $n!(n-1)!$.

RHS = $\frac{(2n-1)!}{(n-1)!n!} + \frac{(2n-1)!}{n!(n-1)!}$

The common denominator is $n!(n-1)!$. Multiply the numerator and denominator of the first fraction by $n$ and the second fraction by $n$. No, just multiply the first fraction by $n/n$ to get $n!$ in the denominator and the second by nothing as it already has $n!$

RHS = $\frac{(2n-1)!}{(n-1)!n!} = \frac{(2n-1)! \times n}{(n-1)! n! \times n}$ No, this gives $n!n!$ denominator.

Let's rewrite the denominators as $n!(n-1)!$ for both terms.

RHS = $\frac{(2n-1)!}{(n-1)!n!} + \frac{(2n-1)!}{n!(n-1)!}$

RHS = $\frac{(2n-1)!}{(n-1)!n!} + \frac{(2n-1)!}{(n-1)!n!}$

RHS = $\frac{(2n-1)! + (2n-1)!}{(n-1)!n!}$

RHS = $\frac{2 \times (2n-1)!}{(n-1)!n!}$

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Now, let's look at the left side of the identity:

LHS = $\binom{2n}{n} = \frac{(2n)!}{n!(2n-n)!} = \frac{(2n)!}{n!n!}$

We know that $(2n)! = 2n \times (2n-1)!$.

LHS = $\frac{2n \times (2n-1)!}{n!n!}$

We know that $n! = n \times (n-1)!$. Substitute this for one of the $n!$ in the denominator:

LHS = $\frac{2n \times (2n-1)!}{n! \times (n \times (n-1)!)}$

Rearrange the denominator: $n! \times n \times (n-1)! = n \times n! \times (n-1)!$. This is not helping.

Let's use $n! = n \times (n-1)!$ on one of the $n!$ in the denominator: $\frac{(2n)!}{n!n!} = \frac{(2n)!}{n! \times n \times (n-1)!}$. This does not simplify to the RHS form.

Let's use the expansion $(2n)! = 2n \times (2n-1)!$ in the numerator of the LHS.

LHS = $\frac{2n \times (2n-1)!}{n!n!}$

Now, let's expand one of the $n!$ in the denominator as $n \times (n-1)!$:

LHS = $\frac{2n \times (2n-1)!}{n! \times n \times (n-1)!}$ No.

LHS = $\frac{2n \times (2n-1)!}{n \times (n-1)! \times n!}$

Cancel the term 'n' from the numerator and denominator:

LHS = $\frac{2 \times (2n-1)!}{(n-1)! \times n!}$

LHS = $\frac{2 \times (2n-1)!}{n!(n-1)!}$

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Comparing the simplified LHS and RHS:

LHS = $\frac{2 \times (2n-1)!}{n!(n-1)!}$

RHS = $\frac{2 \times (2n-1)!}{(n-1)!n!}$

LHS = RHS.

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Thus, we have shown that $\binom{2n}{n} = \binom{2n-1}{n-1} + \binom{2n-1}{n}$.

This proves that the coefficient of the middle term in the expansion of $(1 + x)^{2n}$ is equal to the sum of the coefficients of the two middle terms in the expansion of $(1 + x)^{2n – 1}$.

Hence Showed.

Given expression:

The product $(1 + 2a)^4 (2 – a)^5$

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To Find:

The coefficient of $a^4$ in the expansion of the product.

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Solution:

We will expand each factor using the Binomial Theorem.

The Binomial Theorem states that $(x+y)^n = \sum\limits_{k=0}^{n} \binom{n}{k} x^{n-k} y^k$.

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Consider the first factor, $(1 + 2a)^4$. Here $x=1$, $y=2a$, $n=4$.

The general term in this expansion is $T_{r+1} = \binom{4}{r} (1)^{4-r} (2a)^r = \binom{4}{r} 1 \cdot 2^r a^r = \binom{4}{r} 2^r a^r$.

The terms in the expansion of $(1 + 2a)^4$ are of the form $\binom{4}{r} 2^r a^r$ for $r = 0, 1, 2, 3, 4$.

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Consider the second factor, $(2 – a)^5$. Here $x=2$, $y=-a$, $n=5$.

The general term in this expansion is $T'_{k+1} = \binom{5}{k} (2)^{5-k} (-a)^k = \binom{5}{k} 2^{5-k} (-1)^k a^k$.

The terms in the expansion of $(2 – a)^5$ are of the form $\binom{5}{k} 2^{5-k} (-1)^k a^k$ for $k = 0, 1, 2, 3, 4, 5$.

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The product is $(1 + 2a)^4 (2 – a)^5$. When we multiply the expansions of these two factors, a term containing $a^4$ is obtained by multiplying a term with $a^r$ from the first expansion by a term with $a^k$ from the second expansion such that the sum of the exponents is 4, i.e., $r+k=4$.

The possible pairs of $(r, k)$ such that $r+k=4$, with $0 \leq r \leq 4$ and $0 \leq k \leq 5$, are:

$(r, k) \in \{(0, 4), (1, 3), (2, 2), (3, 1), (4, 0)\}$

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The coefficient of $a^4$ in the product is the sum of the products of the coefficients of the terms $a^r$ and $a^k$ for each of these pairs $(r, k)$.

Coefficient of $a^4 = \sum_{r+k=4} (\text{Coefficient of } a^r \text{ in } (1+2a)^4) \times (\text{Coefficient of } a^k \text{ in } (2-a)^5)$

Coefficient of $a^4 = \binom{4}{0}2^0 \binom{5}{4}2^{5-4}(-1)^4 + \binom{4}{1}2^1 \binom{5}{3}2^{5-3}(-1)^3 + \binom{4}{2}2^2 \binom{5}{2}2^{5-2}(-1)^2 + \binom{4}{3}2^3 \binom{5}{1}2^{5-1}(-1)^1 + \binom{4}{4}2^4 \binom{5}{0}2^{5-0}(-1)^0$

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Let's calculate each part:

For $(r,k)=(0,4)$: $\binom{4}{0}2^0 \times \binom{5}{4}2^{1}(-1)^4 = (1 \times 1) \times (5 \times 2 \times 1) = 1 \times 10 = 10$

For $(r,k)=(1,3)$: $\binom{4}{1}2^1 \times \binom{5}{3}2^{2}(-1)^3 = (4 \times 2) \times (10 \times 4 \times -1) = 8 \times (-40) = -320$

For $(r,k)=(2,2)$: $\binom{4}{2}2^2 \times \binom{5}{2}2^{3}(-1)^2 = (6 \times 4) \times (10 \times 8 \times 1) = 24 \times 80 = 1920$

For $(r,k)=(3,1)$: $\binom{4}{3}2^3 \times \binom{5}{1}2^{4}(-1)^1 = (4 \times 8) \times (5 \times 16 \times -1) = 32 \times (-80) = -2560$

For $(r,k)=(4,0)$: $\binom{4}{4}2^4 \times \binom{5}{0}2^{5}(-1)^0 = (1 \times 16) \times (1 \times 32 \times 1) = 16 \times 32 = 512$

---

The coefficient of $a^4$ is the sum of these values:

$10 + (-320) + 1920 + (-2560) + 512$

$= 10 - 320 + 1920 - 2560 + 512$

$= (10 + 1920 + 512) - (320 + 2560)$

$= 2442 - 2880$

$= -438$

---

The coefficient of $a^4$ in the product $(1 + 2a)^4 (2 – a)^5$ is $\mathbf{-438}$.

Given expansion:

$(x + a)^n$

---

To Find:

The $r^{\text{th}}$ term from the end of the expansion.

---

Solution:

The total number of terms in the expansion of $(x + a)^n$ is $n+1$.

---

Let's consider the terms from the beginning:

1^st term from beginning is $T_1 = \binom{n}{0}x^n a^0$

2^nd term from beginning is $T_2 = \binom{n}{1}x^{n-1} a^1$

$k^{\text{th}}$ term from beginning is $T_k = \binom{n}{k-1}x^{n-(k-1)} a^{k-1}$

$(r+1)^{\text{th}}$ term from beginning is $T_{r+1} = \binom{n}{r}x^{n-r} a^{r}$

---

Now consider the terms from the end:

1^st term from the end is the last term, which is the $(n+1)^{\text{th}}$ term from the beginning.

2^nd term from the end is the $n^{\text{th}}$ term from the beginning.

3^rd term from the end is the $(n-1)^{\text{th}}$ term from the beginning.

The $r^{\text{th}}$ term from the end is the $(n+1 - r + 1)^{\text{th}}$ term from the beginning.

The $r^{\text{th}}$ term from the end is the $(n-r+2)^{\text{th}}$ term from the beginning.

---

To find the $(n-r+2)^{\text{th}}$ term from the beginning using the general term formula $T_{k} = \binom{n}{k-1}x^{n-(k-1)} a^{k-1}$, we set $k = n-r+2$.

The index for the binomial coefficient is $k-1 = (n-r+2) - 1 = n-r+1$.

The exponent of $x$ is $n-(k-1) = n-(n-r+1) = n-n+r-1 = r-1$.

The exponent of $a$ is $k-1 = n-r+1$.

So, the $(n-r+2)^{\text{th}}$ term from the beginning is $T_{n-r+2} = \binom{n}{n-r+1} x^{r-1} a^{n-r+1}$.

---

Alternatively, we can consider the expansion of $(a+x)^n$. The $r^{\text{th}}$ term from the beginning in $(a+x)^n$ is the same as the $r^{\text{th}}$ term from the end in $(x+a)^n$.

The general term in the expansion of $(a+x)^n$ is $T'_{k+1} = \binom{n}{k} a^{n-k} x^k$.

The $r^{\text{th}}$ term from the beginning of $(a+x)^n$ corresponds to $k+1 = r$, so $k = r-1$.

Substituting $k=r-1$ into the general term formula for $(a+x)^n$:

$T'_r = T_{(r-1)+1} = \binom{n}{r-1} a^{n-(r-1)} x^{r-1}$

$T'_r = \binom{n}{r-1} a^{n-r+1} x^{r-1}$

---

The $r^{\text{th}}$ term from the end in the expansion of $(x + a)^n$ is the same as the $r^{\text{th}}$ term from the beginning in the expansion of $(a + x)^n$.

The $r^{\text{th}}$ term from the beginning of $(a+x)^n$ is $T_r = \binom{n}{r-1} a^{n-r+1} x^{r-1}$.

---

Using the property $\binom{n}{k} = \binom{n}{n-k}$, we can see that $\binom{n}{n-r+1} = \binom{n}{n-(n-r+1)} = \binom{n}{n-n+r-1} = \binom{n}{r-1}$.

So the two methods give the same result.

The $r^{\text{th}}$ term from the end in the expansion of $(x + a)^n$ is $\binom{n}{r-1} x^{r-1} a^{n-r+1}$.

---

The $r^{\text{th}}$ term from the end in the expansion of $(x + a)^n$ is $\mathbf{\binom{n}{r-1} x^{r-1} a^{n-r+1}}$.

Given expansion:

$\left( \sqrt [3] {x} + \frac{1}{2\sqrt [3] {x}} \right)^{18}$

---

To Find:

The term independent of $x$ in the expansion.

---

Solution:

The expansion is of the form $(a+b)^n$, where:

$a = \sqrt[3]{x} = x^{1/3}$

$b = \frac{1}{2\sqrt[3]{x}} = \frac{1}{2x^{1/3}} = \frac{1}{2}x^{-1/3}$

$n = 18$

---

The general term (the $(r+1)^{\text{th}}$ term) in the expansion of $(a+b)^n$ is given by the formula:

$T_{r+1} = \binom{n}{r} a^{n-r} b^r$

Substitute the values $n=18$, $a=x^{1/3}$, and $b=\frac{1}{2}x^{-1/3}$:

$T_{r+1} = \binom{18}{r} (x^{1/3})^{18-r} \left(\frac{1}{2}x^{-1/3}\right)^r$

$T_{r+1} = \binom{18}{r} x^{(1/3)(18-r)} \left(\frac{1}{2}\right)^r (x^{-1/3})^r$

$T_{r+1} = \binom{18}{r} x^{\frac{18-r}{3}} \left(\frac{1}{2}\right)^r x^{-\frac{r}{3}}$

Combine the terms with $x$ by adding their exponents:

$T_{r+1} = \binom{18}{r} \left(\frac{1}{2}\right)^r x^{\frac{18-r}{3} - \frac{r}{3}}$

$T_{r+1} = \binom{18}{r} \left(\frac{1}{2}\right)^r x^{\frac{18-2r}{3}}$

---

The term independent of $x$ is the term where the exponent of $x$ is 0.

Set the exponent of $x$ in the general term equal to 0:

$\frac{18-2r}{3} = 0$

$18 - 2r = 0$

$2r = 18$

$r = 9$

---

So, the term independent of $x$ is the $(9+1)^{\text{th}}$, which is the 10th term.

Substitute $r=9$ into the expression for the general term, excluding the $x$ part:

The term independent of $x$ is $T_{10} = \binom{18}{9} \left(\frac{1}{2}\right)^9$.

---

Now, calculate the values:

$\binom{18}{9} = \frac{18!}{9!(18-9)!} = \frac{18!}{9!9!}$

$\binom{18}{9} = \frac{18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9!}{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 9!}$

$\binom{18}{9} = \frac{18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10}{362880}$

$\binom{18}{9} = 48620$

$\left(\frac{1}{2}\right)^9 = \frac{1^9}{2^9} = \frac{1}{512}$

---

Substitute these values into the expression for $T_{10}$:

$T_{10} = 48620 \times \frac{1}{512} = \frac{48620}{512}$

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor. Both are divisible by 4.

$\frac{48620 \div 4}{512 \div 4} = \frac{12155}{128}$

---

The term independent of $x$ is $\mathbf{\frac{12155}{128}}$.

Given expansion:

$\left( x-\frac{3}{x^{2}} \right)^{m}$

This is in the form $(a+b)^n$, where $a = x$, $b = -\frac{3}{x^2}$, and $n = m$.

---

The general term (the $(r+1)^{\text{th}}$ term) in the expansion of $(a+b)^n$ is given by the formula:

$T_{r+1} = \binom{n}{r} a^{n-r} b^r$

Substitute the values $n=m$, $a=x$, and $b=-\frac{3}{x^2}$:

$T_{r+1} = \binom{m}{r} (x)^{m-r} \left(-\frac{3}{x^2}\right)^r$

$T_{r+1} = \binom{m}{r} x^{m-r} (-3)^r \left(\frac{1}{x^2}\right)^r$

$T_{r+1} = \binom{m}{r} (-3)^r x^{m-r} (x^{-2})^r$

$T_{r+1} = \binom{m}{r} (-3)^r x^{m-r} x^{-2r}$

Combine the terms with $x$ by adding their exponents:

$T_{r+1} = \binom{m}{r} (-3)^r x^{m-r-2r}$

$T_{r+1} = \binom{m}{r} (-3)^r x^{m-3r}$

---

The coefficients of the first three terms correspond to $r=0$, $r=1$, and $r=2$ in the general term.

Coefficient of the 1^st term ($r=0$): $\binom{m}{0} (-3)^0 = 1 \times 1 = 1$.

Coefficient of the 2^nd term ($r=1$): $\binom{m}{1} (-3)^1 = m \times (-3) = -3m$.

Coefficient of the 3^rd term ($r=2$): $\binom{m}{2} (-3)^2 = \frac{m(m-1)}{2!} \times 9 = \frac{9m(m-1)}{2}$.

---

The sum of these coefficients is given as 559.

$1 + (-3m) + \frac{9m(m-1)}{2} = 559$

$1 - 3m + \frac{9m^2 - 9m}{2} = 559$

Multiply the entire equation by 2 to eliminate the fraction:

$2(1) - 2(3m) + 2\left(\frac{9m^2 - 9m}{2}\right) = 2(559)$

$2 - 6m + 9m^2 - 9m = 1118$

$9m^2 - 15m + 2 = 1118$

Subtract 1118 from both sides to form a quadratic equation:

$9m^2 - 15m + 2 - 1118 = 0$

$9m^2 - 15m - 1116 = 0$

Divide the equation by the greatest common divisor of the coefficients (3):

$\frac{9m^2}{3} - \frac{15m}{3} - \frac{1116}{3} = \frac{0}{3}$

$3m^2 - 5m - 372 = 0$

---

Solve this quadratic equation for $m$ using the quadratic formula: $m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Here, $a=3$, $b=-5$, $c=-372$.

$m = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(3)(-372)}}{2(3)}$

$m = \frac{5 \pm \sqrt{25 + 12 \times 372}}{6}$

$12 \times 372 = 4464$.

$m = \frac{5 \pm \sqrt{25 + 4464}}{6}$

$m = \frac{5 \pm \sqrt{4489}}{6}$

We need to find the square root of 4489. $60^2=3600$, $70^2=4900$. The number ends in 9, so the unit digit of the square root must be 3 or 7. Let's try 67. $67 \times 67 = 4489$.

$m = \frac{5 \pm 67}{6}$

---

We have two possible values for $m$:

$m_1 = \frac{5 + 67}{6} = \frac{72}{6} = 12$

$m_2 = \frac{5 - 67}{6} = \frac{-62}{6} = -\frac{31}{3}$

---

Since $m$ is a natural number, we must choose the positive integer value.

$m = 12$.

---

Now that we have found $m=12$, we need to find the term of the expansion containing $x^3$.

The general term is $T_{r+1} = \binom{m}{r} (-3)^r x^{m-3r}$. Substitute $m=12$:

$T_{r+1} = \binom{12}{r} (-3)^r x^{12-3r}$

We want the term containing $x^3$, so the exponent of $x$ must be 3.

Set the exponent of $x$ equal to 3:

$12 - 3r = 3$

$12 - 3 = 3r$

$9 = 3r$

$r = \frac{9}{3} = 3$

---

So, the term containing $x^3$ is the $(3+1)^{\text{th}}$, which is the 4th term.

Substitute $r=3$ into the general term expression (with $m=12$):

$T_4 = T_{3+1} = \binom{12}{3} (-3)^3 x^{12-3(3)}$

$T_4 = \binom{12}{3} (-3)^3 x^{12-9}$

$T_4 = \binom{12}{3} (-3)^3 x^3$

---

Now, calculate the values of the binomial coefficient and the power of -3:

$\binom{12}{3} = \frac{12!}{3!(12-3)!} = \frac{12!}{3!9!} = \frac{12 \times 11 \times 10 \times 9!}{3 \times 2 \times 1 \times 9!} = \frac{12 \times 11 \times 10}{6} = 2 \times 11 \times 10 = 220$

$(-3)^3 = (-1)^3 \times 3^3 = -1 \times 27 = -27$

---

Substitute these values back into the expression for $T_4$:

$T_4 = 220 \times (-27) \times x^3$

Calculate $220 \times (-27)$: $220 \times 27 = 5940$. So, $220 \times (-27) = -5940$.

---

The term containing $x^3$ is $-5940 x^3$.

The term of the expansion containing $x^3$ is $\mathbf{-5940 x^3}$.

Given expansion:

$(1 + x)^{34}$

This is in the form $(1+a)^n$, where $a=x$ and $n=34$.

---

To Find:

The value of $r$ such that the coefficients of the $(r-5)^{\text{th}}$ term and the $(2r-1)^{\text{th}}$ term are equal.

---

Solution:

The general term in the expansion of $(1+x)^n$ is the $(k+1)^{\text{th}}$ term, given by $T_{k+1} = \binom{n}{k} 1^{n-k} x^k = \binom{n}{k} x^k$.

The coefficient of the $(k+1)^{\text{th}}$ term is $\binom{n}{k}$.

---

For the $(r-5)^{\text{th}}$ term, the term number is $r-5$. So, $k+1 = r-5$, which means $k = r-5-1 = r-6$.

The coefficient of the $(r-5)^{\text{th}}$ term is $\binom{34}{r-6}$.

For this coefficient to be well-defined, the lower index must be a non-negative integer and less than or equal to the upper index. So, $r-6 \geq 0$ and $r-6 \leq 34$.

$r \geq 6$ and $r \leq 40$.

---

For the $(2r-1)^{\text{th}}$ term, the term number is $2r-1$. So, $k+1 = 2r-1$, which means $k = 2r-1-1 = 2r-2$.

The coefficient of the $(2r-1)^{\text{th}}$ term is $\binom{34}{2r-2}$.

For this coefficient to be well-defined, the lower index must be a non-negative integer and less than or equal to the upper index. So, $2r-2 \geq 0$ and $2r-2 \leq 34$.

$2r \geq 2 \implies r \geq 1$.

$2r \leq 36 \implies r \leq 18$.

---

We are given that the coefficients of the $(r-5)^{\text{th}}$ term and the $(2r-1)^{\text{th}}$ term are equal.

$\binom{34}{r-6} = \binom{34}{2r-2}$

---

We use the property of binomial coefficients: If $\binom{n}{a} = \binom{n}{b}$, then either $a=b$ or $a=n-b$.

In this case, $n=34$, $a=r-6$, and $b=2r-2$.

---

Case 1: The lower indices are equal.

$r-6 = 2r-2$

Rearranging the terms:

$-6 + 2 = 2r - r$

$-4 = r$

$r = -4$

---

Case 2: The sum of the lower indices equals the upper index.

$(r-6) + (2r-2) = 34$

$r - 6 + 2r - 2 = 34$

$3r - 8 = 34$

$3r = 34 + 8$

$3r = 42$

$r = \frac{42}{3} = 14$

---

Now we check if these values of $r$ satisfy the validity conditions for the indices derived earlier ($r \geq 6$, $r \leq 40$, $r \geq 1$, $r \leq 18$). The combined conditions are $6 \leq r \leq 18$.

For $r = -4$:

$r \geq 6$ ($-4 \geq 6$) is False.

This value of $r$ is not valid as the $(r-5)^{\text{th}}$ term would be the $(-9)^{\text{th}}$ term, which does not exist in a standard binomial expansion starting from the 1st term.

For $r = 14$:

$r \geq 6$ ($14 \geq 6$) is True.

$r \leq 40$ ($14 \leq 40$) is True.

$r \geq 1$ ($14 \geq 1$) is True.

$r \leq 18$ ($14 \leq 18$) is True.

All conditions are satisfied for $r=14$. The terms are the $(14-5)^{\text{th}} = 9^{\text{th}}$ term (coefficient $\binom{34}{8}$) and the $(2 \times 14 - 1)^{\text{th}} = (28-1)^{\text{th}} = 27^{\text{th}}$ term (coefficient $\binom{34}{26}$).

Check $\binom{34}{8} = \binom{34}{26}$: This is true because $\binom{n}{k} = \binom{n}{n-k}$, so $\binom{34}{8} = \binom{34}{34-8} = \binom{34}{26}$.

---

The only valid value for $r$ is 14.

The value of r is $\mathbf{14}$.

Given:

The first three terms of the expansion $(a + b)^n$ are 729, 7290, and 30375.

Let the terms be $T_1$, $T_2$, and $T_3$.

$T_1 = 729$

$T_2 = 7290$

$T_3 = 30375$

---

To Find:

The values of $a$, $b$, and $n$.

---

Solution:

The general term $(k+1)^{\text{th}}$ in the expansion of $(a+b)^n$ is given by $T_{k+1} = \binom{n}{k} a^{n-k} b^k$.

The first term is $T_1$ (when $k=0$):

$T_1 = \binom{n}{0} a^{n-0} b^0 = 1 \cdot a^n \cdot 1 = a^n$

---

The second term is $T_2$ (when $k=1$):

$T_2 = \binom{n}{1} a^{n-1} b^1 = n a^{n-1} b$

---

The third term is $T_3$ (when $k=2$):

$T_3 = \binom{n}{2} a^{n-2} b^2 = \frac{n(n-1)}{2} a^{n-2} b^2$

---

Divide the second term by the first term:

$\frac{T_2}{T_1} = \frac{n a^{n-1} b}{a^n} = \frac{7290}{729}$

$n \frac{a^{n-1}}{a^n} b = 10$

$n a^{n-1-n} b = 10$

$n a^{-1} b = 10$

---

Divide the third term by the second term:

$\frac{T_3}{T_2} = \frac{\frac{n(n-1)}{2} a^{n-2} b^2}{n a^{n-1} b} = \frac{30375}{7290}$

Simplify the numerical fraction:

$\frac{30375}{7290} = \frac{\cancel{30375}^{6075}}{\cancel{7290}_{1458}} = \frac{\cancel{6075}^{675}}{\cancel{1458}_{162}} = \frac{\cancel{675}^{75}}{\cancel{162}_{18}} = \frac{\cancel{75}^{25}}{\cancel{18}_{6}} = \frac{25}{6}$

So, $\frac{\frac{n(n-1)}{2} a^{n-2} b^2}{n a^{n-1} b} = \frac{25}{6}$

$\frac{n(n-1)}{2n} \frac{a^{n-2}}{a^{n-1}} \frac{b^2}{b} = \frac{25}{6}$

$\frac{n-1}{2} a^{(n-2)-(n-1)} b^{2-1} = \frac{25}{6}$

$\frac{n-1}{2} a^{-1} b^1 = \frac{25}{6}$

---

From equation (iv), we have $\frac{b}{a} = \frac{10}{n}$. Substitute this into equation (v):

$\frac{n-1}{2} \left(\frac{10}{n}\right) = \frac{25}{6}$

$\frac{10(n-1)}{2n} = \frac{25}{6}$

$\frac{5(n-1)}{n} = \frac{25}{6}$

Divide both sides by 5:

$\frac{n-1}{n} = \frac{5}{6}$

Cross-multiply:

$6(n-1) = 5n$

$6n - 6 = 5n$

$6n - 5n = 6$

---

Substitute $n=6$ into equation (iv):

$6 \frac{b}{a} = 10$

---

Substitute $n=6$ into equation (i):

$a^n = 729$

$a^6 = 729$

Since $3^6 = (3^2)^3 = 9^3 = 729$, we have $a=3$ (assuming $a$ is real and positive, typical for expansion terms). If $a$ could be negative, $a=-3$ is also possible, but usually in such problems $a$ and $b$ are assumed positive unless specified otherwise).

---

Substitute $a=3$ into equation (vii):

$\frac{b}{3} = \frac{5}{3}$

$b = \frac{5}{3} \times 3$

---

Verify the values with the original terms:

$T_1 = a^n = 3^6 = 729$. Correct.

$T_2 = na^{n-1}b = 6 \times 3^{6-1} \times 5 = 6 \times 3^5 \times 5 = 6 \times 243 \times 5 = 30 \times 243 = 7290$. Correct.

$T_3 = \frac{n(n-1)}{2} a^{n-2} b^2 = \frac{6(6-1)}{2} \times 3^{6-2} \times 5^2 = \frac{6 \times 5}{2} \times 3^4 \times 25 = 15 \times 81 \times 25 = 15 \times 2025 = 30375$. Correct.

---

The values are $\mathbf{a=3}$, $\mathbf{b=5}$, and $\mathbf{n=6}$.

Given expansion:

$(3 + ax)^9$

This is in the form $(u+v)^n$, where $u=3$, $v=ax$, and $n=9$.

---

To Find:

The value of $a$ such that the coefficients of $x^2$ and $x^3$ in the expansion are equal.

---

Solution:

The general term (the $(r+1)^{\text{th}}$ term) in the expansion of $(u+v)^n$ is given by the formula:

$T_{r+1} = \binom{n}{r} u^{n-r} v^r$

Substitute the values $n=9$, $u=3$, and $v=ax$:

$T_{r+1} = \binom{9}{r} (3)^{9-r} (ax)^r$

$T_{r+1} = \binom{9}{r} 3^{9-r} a^r x^r$

$T_{r+1} = \binom{9}{r} 3^{9-r} a^r x^r$

The coefficient of the term containing $x^r$ is $\binom{9}{r} 3^{9-r} a^r$.

---

We are looking for the coefficients of $x^2$ and $x^3$.

For the coefficient of $x^2$, we set the exponent of $x$ equal to 2, so $r=2$.

Coefficient of $x^2 = \binom{9}{2} 3^{9-2} a^2 = \binom{9}{2} 3^7 a^2$.

For the coefficient of $x^3$, we set the exponent of $x$ equal to 3, so $r=3$.

Coefficient of $x^3 = \binom{9}{3} 3^{9-3} a^3 = \binom{9}{3} 3^6 a^3$.

---

We are given that these coefficients are equal.

Coefficient of $x^2$ = Coefficient of $x^3$

---

Calculate the binomial coefficients:

$\binom{9}{2} = \frac{9!}{2!(9-2)!} = \frac{9!}{2!7!} = \frac{9 \times 8}{2 \times 1} = 36$.

$\binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84$.

---

Substitute these values into equation (i):

$36 \times 3^7 \times a^2 = 84 \times 3^6 \times a^3$

---

We assume $a \neq 0$, otherwise the coefficients of $x^2$ and $x^3$ would be 0 (for $r \geq 1$), but the coefficient of $x^0$ is non-zero, so the terms exist. If $a=0$, the expansion is just $3^9$, the only non-zero coefficient is for $x^0$, which is 1. The coefficients of $x^2$ and $x^3$ would be 0, and $0=0$, so $a=0$ is a possible solution if the problem statement allows $a=0$. However, usually problems imply the existence of the terms with non-zero coefficients unless $n$ is too small. Here $n=9$, so terms with $x^2$ and $x^3$ exist and have non-zero combinatorial factors.

Divide both sides by $3^6 a^2$ (assuming $a \neq 0$ and $3^6 \neq 0$):

$\frac{36 \times 3^7 \times a^2}{3^6 a^2} = \frac{84 \times 3^6 \times a^3}{3^6 a^2}$

$36 \times 3^{7-6} \times a^{2-2} = 84 \times 3^{6-6} \times a^{3-2}$

$36 \times 3^1 \times a^0 = 84 \times 3^0 \times a^1$

$36 \times 3 \times 1 = 84 \times 1 \times a$

$108 = 84a$

$a = \frac{108}{84}$

---

Simplify the fraction $\frac{108}{84}$. Both numerator and denominator are divisible by 12.

$\frac{108 \div 12}{84 \div 12} = \frac{9}{7}$

$a = \frac{9}{7}$

---

The value of $a$ is $\mathbf{\frac{9}{7}}$.

Given expression:

The product $(1 + 2x)^6 (1 – x)^7$

---

To Find:

The coefficient of $x^5$ in the expansion of the product.

---

Solution:

We will expand each factor using the Binomial Theorem.

The Binomial Theorem states that $(u+v)^n = \sum\limits_{k=0}^{n} \binom{n}{k} u^{n-k} v^k$. For $(1+y)^n$, the general term is $\binom{n}{k} y^k$.

---

Consider the first factor, $(1 + 2x)^6$. Here $u=1$, $v=2x$, $n=6$.

The general term in this expansion is $T_{r+1} = \binom{6}{r} (1)^{6-r} (2x)^r = \binom{6}{r} 1 \cdot 2^r x^r = \binom{6}{r} 2^r x^r$.

The terms in the expansion of $(1 + 2x)^6$ are of the form $\binom{6}{r} 2^r x^r$ for $r = 0, 1, 2, 3, 4, 5, 6$.

---

Consider the second factor, $(1 – x)^7$. Here $u=1$, $v=-x$, $n=7$.

The general term in this expansion is $T'_{k+1} = \binom{7}{k} (1)^{7-k} (-x)^k = \binom{7}{k} 1 \cdot (-1)^k x^k = \binom{7}{k} (-1)^k x^k$.

The terms in the expansion of $(1 – x)^7$ are of the form $\binom{7}{k} (-1)^k x^k$ for $k = 0, 1, 2, 3, 4, 5, 6, 7$.

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The product is $(1 + 2x)^6 (1 – x)^7$. When we multiply the expansions of these two factors, a term containing $x^5$ is obtained by multiplying a term with $x^r$ from the first expansion by a term with $x^k$ from the second expansion such that the sum of the exponents is 5, i.e., $r+k=5$.

The possible pairs of $(r, k)$ such that $r+k=5$, with $0 \leq r \leq 6$ and $0 \leq k \leq 7$, are:

$(r, k) \in \{(0, 5), (1, 4), (2, 3), (3, 2), (4, 1), (5, 0)\}$

---

The coefficient of $x^5$ in the product is the sum of the products of the coefficients of the terms $x^r$ and $x^k$ for each of these pairs $(r, k)$.

Coefficient of $x^5 = \sum_{r+k=5} (\text{Coefficient of } x^r \text{ in } (1+2x)^6) \times (\text{Coefficient of } x^k \text{ in } (1-x)^7)$

Coefficient of $x^5 = \binom{6}{0}2^0 \binom{7}{5}(-1)^5 + \binom{6}{1}2^1 \binom{7}{4}(-1)^4 + \binom{6}{2}2^2 \binom{7}{3}(-1)^3 + \binom{6}{3}2^3 \binom{7}{2}(-1)^2 + \binom{6}{4}2^4 \binom{7}{1}(-1)^1 + \binom{6}{5}2^5 \binom{7}{0}(-1)^0$

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Let's calculate each part:

For $(r,k)=(0,5)$: $\binom{6}{0}2^0 \times \binom{7}{5}(-1)^5 = (1 \times 1) \times (\frac{7 \times 6}{2} \times -1) = 1 \times (21 \times -1) = -21$

For $(r,k)=(1,4)$: $\binom{6}{1}2^1 \times \binom{7}{4}(-1)^4 = (6 \times 2) \times (\frac{7 \times 6 \times 5}{3 \times 2 \times 1} \times 1) = 12 \times (35 \times 1) = 420$

For $(r,k)=(2,3)$: $\binom{6}{2}2^2 \times \binom{7}{3}(-1)^3 = (\frac{6 \times 5}{2} \times 4) \times (\frac{7 \times 6 \times 5}{3 \times 2 \times 1} \times -1) = (15 \times 4) \times (35 \times -1) = 60 \times (-35) = -2100$

For $(r,k)=(3,2)$: $\binom{6}{3}2^3 \times \binom{7}{2}(-1)^2 = (\frac{6 \times 5 \times 4}{3 \times 2 \times 1} \times 8) \times (\frac{7 \times 6}{2 \times 1} \times 1) = (20 \times 8) \times (21 \times 1) = 160 \times 21 = 3360$

For $(r,k)=(4,1)$: $\binom{6}{4}2^4 \times \binom{7}{1}(-1)^1 = (\frac{6 \times 5}{2} \times 16) \times (7 \times -1) = (15 \times 16) \times (-7) = 240 \times (-7) = -1680$

For $(r,k)=(5,0)$: $\binom{6}{5}2^5 \times \binom{7}{0}(-1)^0 = (6 \times 32) \times (1 \times 1) = 192 \times 1 = 192$

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The coefficient of $x^5$ is the sum of these values:

$-21 + 420 + (-2100) + 3360 + (-1680) + 192$

$= -21 + 420 - 2100 + 3360 - 1680 + 192$

$= (420 + 3360 + 192) - (21 + 2100 + 1680)$

$= 3972 - 3801$

$= 171$

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The coefficient of $x^5$ in the product $(1 + 2x)^6 (1 – x)^7$ is $\mathbf{171}$.

Given:

$a$ and $b$ are distinct integers.

$n$ is a positive integer.

---

To Prove:

$a - b$ is a factor of $a^n - b^n$.

This is equivalent to proving that $a^n - b^n$ is divisible by $a-b$.

---

Proof (using the hint):

Write $a^n$ as $(a - b + b)^n$. Let $a-b = x$. Then $a^n = (x+b)^n$.

Expand $(x+b)^n$ using the Binomial Theorem:

$(x+b)^n = \binom{n}{0}x^n b^0 + \binom{n}{1}x^{n-1} b^1 + \binom{n}{2}x^{n-2} b^2 + ... + \binom{n}{n-1}x^1 b^{n-1} + \binom{n}{n}x^0 b^n$

---

Substitute $x = a-b$ back into the expansion:

$a^n = \binom{n}{0}(a-b)^n b^0 + \binom{n}{1}(a-b)^{n-1} b^1 + \binom{n}{2}(a-b)^{n-2} b^2 + ... + \binom{n}{n-1}(a-b)^1 b^{n-1} + \binom{n}{n}(a-b)^0 b^n$

---

Simplify the terms:

$a^n = 1 \cdot (a-b)^n \cdot 1 + n (a-b)^{n-1} b + \binom{n}{2}(a-b)^{n-2} b^2 + ... + n (a-b) b^{n-1} + 1 \cdot 1 \cdot b^n$

$a^n = (a-b)^n + n (a-b)^{n-1} b + \binom{n}{2}(a-b)^{n-2} b^2 + ... + n (a-b) b^{n-1} + b^n$

---

Rearrange the equation to isolate $a^n - b^n$:

$a^n - b^n = (a-b)^n + n (a-b)^{n-1} b + \binom{n}{2}(a-b)^{n-2} b^2 + ... + n (a-b) b^{n-1}$

---

Notice that every term on the right side of the equation has a factor of $(a-b)$ (since $n$ is a positive integer, the lowest power of $(a-b)$ is 1 in the second to last term, and $(a-b)^k$ for $k \geq 1$ is divisible by $(a-b)$).

We can factor out $(a-b)$ from the entire right side:

$a^n - b^n = (a-b) \left[ (a-b)^{n-1} + n (a-b)^{n-2} b + \binom{n}{2}(a-b)^{n-3} b^2 + ... + n b^{n-1} \right]$

Let $K = \left[ (a-b)^{n-1} + n (a-b)^{n-2} b + \binom{n}{2}(a-b)^{n-3} b^2 + ... + n b^{n-1} \right]$.

Since $n$ is a positive integer, $a$ and $b$ are integers, all the terms inside the bracket are integers (assuming $n \geq 1$. If $n=1$, the term is $b^0=1$, which is an integer). Thus, $K$ is an integer.

So, we have $a^n - b^n = (a-b) K$, where $K$ is an integer.

---

This shows that $a^n - b^n$ can be expressed as a multiple of $a-b$. By definition, this means $a^n - b^n$ is divisible by $a-b$.

Therefore, $a - b$ is a factor of $a^n - b^n$ whenever $n$ is a positive integer.

Hence Proved.

Given expression:

$(\sqrt{3} + \sqrt{2})^6 - (\sqrt{3} - \sqrt{2})^6$

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This expression is in the form $(a+b)^n - (a-b)^n$, where $a = \sqrt{3}$, $b = \sqrt{2}$, and $n = 6$.

---

Let's expand $(a+b)^6$ and $(a-b)^6$ using the Binomial Theorem:

$(a+b)^6 = \binom{6}{0}a^6 b^0 + \binom{6}{1}a^5 b^1 + \binom{6}{2}a^4 b^2 + \binom{6}{3}a^3 b^3 + \binom{6}{4}a^2 b^4 + \binom{6}{5}a^1 b^5 + \binom{6}{6}a^0 b^6$

$(a-b)^6 = \binom{6}{0}a^6 (-b)^0 + \binom{6}{1}a^5 (-b)^1 + \binom{6}{2}a^4 (-b)^2 + \binom{6}{3}a^3 (-b)^3 + \binom{6}{4}a^2 (-b)^4 + \binom{6}{5}a^1 (-b)^5 + \binom{6}{6}a^0 (-b)^6$

$(a-b)^6 = \binom{6}{0}a^6 b^0 - \binom{6}{1}a^5 b^1 + \binom{6}{2}a^4 b^2 - \binom{6}{3}a^3 b^3 + \binom{6}{4}a^2 b^4 - \binom{6}{5}a^1 b^5 + \binom{6}{6}a^0 b^6$

---

Now, subtract $(a-b)^6$ from $(a+b)^6$:

$(a+b)^6 - (a-b)^6 = \left(\binom{6}{0}a^6 + \binom{6}{1}a^5 b + \binom{6}{2}a^4 b^2 + \binom{6}{3}a^3 b^3 + \binom{6}{4}a^2 b^4 + \binom{6}{5}a^1 b^5 + \binom{6}{6}a^0 b^6\right) - \left(\binom{6}{0}a^6 - \binom{6}{1}a^5 b + \binom{6}{2}a^4 b^2 - \binom{6}{3}a^3 b^3 + \binom{6}{4}a^2 b^4 - \binom{6}{5}a^1 b^5 + \binom{6}{6}a^0 b^6\right)$

When subtracting, the terms with even powers of $b$ (which are positive in both expansions) will cancel out, and the terms with odd powers of $b$ (which have opposite signs) will be added.

$(a+b)^6 - (a-b)^6 = 2\left[\binom{6}{1}a^5 b^1 + \binom{6}{3}a^3 b^3 + \binom{6}{5}a^1 b^5\right]$

---

Calculate the binomial coefficients:

$\binom{6}{1} = 6$

$\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$

$\binom{6}{5} = \binom{6}{6-5} = \binom{6}{1} = 6$

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Substitute these coefficients and $a=\sqrt{3}$, $b=\sqrt{2}$ into the expression:

$2\left[6(\sqrt{3})^5 (\sqrt{2})^1 + 20(\sqrt{3})^3 (\sqrt{2})^3 + 6(\sqrt{3})^1 (\sqrt{2})^5\right]$

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Calculate the powers of $\sqrt{3}$ and $\sqrt{2}$:

$(\sqrt{3})^1 = \sqrt{3}$

$(\sqrt{3})^3 = (\sqrt{3})^2 \times \sqrt{3} = 3\sqrt{3}$

$(\sqrt{3})^5 = (\sqrt{3})^2 \times (\sqrt{3})^2 \times \sqrt{3} = 3 \times 3 \times \sqrt{3} = 9\sqrt{3}$

$(\sqrt{2})^1 = \sqrt{2}$

$(\sqrt{2})^3 = (\sqrt{2})^2 \times \sqrt{2} = 2\sqrt{2}$

$(\sqrt{2})^5 = (\sqrt{2})^2 \times (\sqrt{2})^2 \times \sqrt{2} = 2 \times 2 \times \sqrt{2} = 4\sqrt{2}$

---

Substitute these values back into the expression:

$2\left[6(9\sqrt{3})(\sqrt{2}) + 20(3\sqrt{3})(2\sqrt{2}) + 6(\sqrt{3})(4\sqrt{2})\right]$

Use the property $\sqrt{u}\sqrt{v} = \sqrt{uv}$:

$2\left[54\sqrt{6} + 20(6\sqrt{6}) + 24\sqrt{6}\right]$

$2\left[54\sqrt{6} + 120\sqrt{6} + 24\sqrt{6}\right]$

Combine the terms with $\sqrt{6}$:

$2\left[(54 + 120 + 24)\sqrt{6}\right]$

$2\left[198\sqrt{6}\right]$

$2 \times 198 \times \sqrt{6} = 396\sqrt{6}$

---

The evaluated value is $\mathbf{396\sqrt{6}}$.

Given expression:

$(a^2 + \sqrt{a^2 - 1})^4 + (a^2 - \sqrt{a^2 - 1})^4$

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This expression is in the form $(x+y)^n + (x-y)^n$, where $x = a^2$, $y = \sqrt{a^2 - 1}$, and $n = 4$.

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Let's expand $(x+y)^4$ and $(x-y)^4$ using the Binomial Theorem:

$(x+y)^4 = \binom{4}{0}x^4 y^0 + \binom{4}{1}x^3 y^1 + \binom{4}{2}x^2 y^2 + \binom{4}{3}x^1 y^3 + \binom{4}{4}x^0 y^4$

$(x-y)^4 = \binom{4}{0}x^4 (-y)^0 + \binom{4}{1}x^3 (-y)^1 + \binom{4}{2}x^2 (-y)^2 + \binom{4}{3}x^1 (-y)^3 + \binom{4}{4}x^0 (-y)^4$

$(x-y)^4 = \binom{4}{0}x^4 y^0 - \binom{4}{1}x^3 y^1 + \binom{4}{2}x^2 y^2 - \binom{4}{3}x^1 y^3 + \binom{4}{4}x^0 y^4$

---

Now, add $(x+y)^4$ and $(x-y)^4$:

$(x+y)^4 + (x-y)^4 = \left(\binom{4}{0}x^4 y^0 + \binom{4}{1}x^3 y^1 + \binom{4}{2}x^2 y^2 + \binom{4}{3}x^1 y^3 + \binom{4}{4}x^0 y^4\right) + \left(\binom{4}{0}x^4 y^0 - \binom{4}{1}x^3 y^1 + \binom{4}{2}x^2 y^2 - \binom{4}{3}x^1 y^3 + \binom{4}{4}x^0 y^4\right)$

When adding, the terms with odd powers of $y$ (which have opposite signs) will cancel out, and the terms with even powers of $y$ (which have the same sign) will be added.

$(x+y)^4 + (x-y)^4 = 2\left[\binom{4}{0}x^4 y^0 + \binom{4}{2}x^2 y^2 + \binom{4}{4}x^0 y^4\right]$

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Calculate the binomial coefficients:

$\binom{4}{0} = 1$

$\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$

$\binom{4}{4} = 1$

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Substitute these coefficients and $x = a^2$, $y = \sqrt{a^2 - 1}$ into the expression:

$2\left[1 \cdot (a^2)^4 (\sqrt{a^2 - 1})^0 + 6 \cdot (a^2)^2 (\sqrt{a^2 - 1})^2 + 1 \cdot (a^2)^0 (\sqrt{a^2 - 1})^4\right]$

$2\left[1 \cdot a^8 \cdot 1 + 6 \cdot a^4 \cdot (a^2 - 1) + 1 \cdot 1 \cdot (a^2 - 1)^2\right]$

---

Expand and simplify the terms inside the bracket:

$a^8 + 6a^4(a^2 - 1) + (a^2 - 1)^2$

$a^8 + 6a^6 - 6a^4 + (a^4 - 2a^2 + 1)$

$a^8 + 6a^6 - 6a^4 + a^4 - 2a^2 + 1$

$a^8 + 6a^6 + (-6a^4 + a^4) - 2a^2 + 1$

$a^8 + 6a^6 - 5a^4 - 2a^2 + 1$

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Multiply by 2:

$2(a^8 + 6a^6 - 5a^4 - 2a^2 + 1)$

$2a^8 + 12a^6 - 10a^4 - 4a^2 + 2$

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The value of the expression is $\mathbf{2a^8 + 12a^6 - 10a^4 - 4a^2 + 2}$.

Given expression:

$(0.99)^5$

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To Find:

An approximation of $(0.99)^5$ using the first three terms of its binomial expansion.

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Solution:

We can write $0.99$ as $1 - 0.01$. So, we need to consider the expansion of $(1 - 0.01)^5$.

This is in the form $(a+b)^n$, where $a = 1$, $b = -0.01$, and $n = 5$.

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We use the Binomial Theorem for expansion:

$(a+b)^n = \binom{n}{0}a^n b^0 + \binom{n}{1}a^{n-1} b^1 + \binom{n}{2}a^{n-2} b^2 + \binom{n}{3}a^{n-3} b^3 + ... + \binom{n}{n}a^0 b^n$

For $(1 - 0.01)^5$, the first three terms correspond to $r=0$, $r=1$, and $r=2$ in the general term $T_{r+1} = \binom{5}{r} (1)^{5-r} (-0.01)^r = \binom{5}{r} (-0.01)^r$.

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First term ($r=0$): $T_1 = \binom{5}{0} (-0.01)^0 = 1 \times 1 = 1$.

Second term ($r=1$): $T_2 = \binom{5}{1} (-0.01)^1 = 5 \times (-0.01) = -0.05$.

Third term ($r=2$): $T_3 = \binom{5}{2} (-0.01)^2 = \frac{5 \times 4}{2 \times 1} \times (0.0001) = 10 \times 0.0001 = 0.0010$.

---

The approximation using the first three terms is the sum of these three terms:

$(0.99)^5 \approx T_1 + T_2 + T_3$

$(0.99)^5 \approx 1 + (-0.05) + 0.0010$

$(0.99)^5 \approx 1 - 0.05 + 0.0010$

$(0.99)^5 \approx 0.95 + 0.0010$

$(0.99)^5 \approx 0.9510$

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The approximation of $(0.99)^5$ using the first three terms is $\mathbf{0.9510}$.

Given expansion:

$\left( \sqrt [4] {2} + \frac{1}{\sqrt [4] {3}} \right)^n$

This is in the form $(a+b)^n$, where $a = \sqrt[4]{2} = 2^{1/4}$ and $b = \frac{1}{\sqrt[4]{3}} = 3^{-1/4}$.

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To Find:

The value of $n$ such that the ratio of the fifth term from the beginning to the fifth term from the end is $\sqrt{6} : 1$.

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Solution:

The general term (the $(k+1)^{\text{th}}$ term) in the expansion of $(a+b)^n$ is given by the formula:

$T_{k+1} = \binom{n}{k} a^{n-k} b^k$

---

The fifth term from the beginning corresponds to $k+1 = 5$, so $k=4$.

$T_5 (\text{from beginning}) = T_{4+1} = \binom{n}{4} a^{n-4} b^4$

Substituting $a=2^{1/4}$ and $b=3^{-1/4}$:

$T_5 (\text{from beginning}) = \binom{n}{4} (2^{1/4})^{n-4} (3^{-1/4})^4 = \binom{n}{4} 2^{(n-4)/4} 3^{-4/4} = \binom{n}{4} 2^{(n-4)/4} 3^{-1}$

---

The total number of terms in the expansion is $n+1$.

The $r^{\text{th}}$ term from the end is the $(n+1 - r + 1)^{\text{th}} = (n-r+2)^{\text{th}}$ term from the beginning.

The fifth term from the end corresponds to $r=5$. Its position from the beginning is $(n-5+2)^{\text{th}} = (n-3)^{\text{th}}$.

So, the fifth term from the end is $T_{n-3}$ from the beginning. This corresponds to $k+1 = n-3$, so $k = n-4$.

$T_5 (\text{from end}) = T_{n-4+1} = \binom{n}{n-4} a^{n-(n-4)} b^{n-4} = \binom{n}{n-4} a^4 b^{n-4}$

Substituting $a=2^{1/4}$ and $b=3^{-1/4}$:

$T_5 (\text{from end}) = \binom{n}{n-4} (2^{1/4})^4 (3^{-1/4})^{n-4} = \binom{n}{n-4} 2^{4/4} 3^{-(n-4)/4} = \binom{n}{n-4} 2^1 3^{-(n-4)/4}$

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We are given that the ratio of the fifth term from the beginning to the fifth term from the end is $\sqrt{6} : 1$.

$\frac{T_5 (\text{from beginning})}{T_5 (\text{from end})} = \frac{\sqrt{6}}{1} = \sqrt{6}$

$\frac{\binom{n}{4} 2^{(n-4)/4} 3^{-1}}{\binom{n}{n-4} 2^1 3^{-(n-4)/4}} = \sqrt{6}$

Using the property $\binom{n}{k} = \binom{n}{n-k}$, we have $\binom{n}{4} = \binom{n}{n-4}$. These terms cancel out.

$\frac{2^{(n-4)/4} 3^{-1}}{2^1 3^{-(n-4)/4}} = \sqrt{6}$

Separate the terms with base 2 and 3:

$\frac{2^{(n-4)/4}}{2^1} \times \frac{3^{-1}}{3^{-(n-4)/4}} = \sqrt{6}$

Using the exponent rule $\frac{x^p}{x^q} = x^{p-q}$ and $\frac{1}{x^q} = x^{-q}$:

$2^{((n-4)/4) - 1} \times 3^{-1 - (-(n-4)/4)} = \sqrt{6}$

$2^{(n-4-4)/4} \times 3^{-1 + (n-4)/4} = \sqrt{6}$

$2^{(n-8)/4} \times 3^{(n-4-4)/4} = \sqrt{6}$

$2^{(n-8)/4} \times 3^{(n-8)/4} = \sqrt{6}$

Using the property $(uv)^p = u^p v^p$ in reverse:

$(2 \times 3)^{(n-8)/4} = \sqrt{6}$

$6^{(n-8)/4} = 6^{1/2}$

---

Since the bases are equal, the exponents must be equal:

$\frac{n-8}{4} = \frac{1}{2}$

Multiply both sides by 4:

$n-8 = \frac{1}{2} \times 4$

$n-8 = 2$

$n = 8 + 2$

$n = 10$

---

We should verify if $n=10$ is valid. The 5th term from the beginning exists if $n \geq 4$, and the 5th term from the end (which is the $10-5+2=7$th term from the beginning) exists if $n \geq 6$. Both conditions are satisfied for $n=10$.

---

The value of n is $\mathbf{10}$.

Given expression:

$\left(1+ \frac{x}{2} -\frac{2}{x}\right)^{4}$

---

To Expand:

Expand the given trinomial using the Binomial Theorem.

---

Solution:

The Binomial Theorem applies to binomials (expressions with two terms). We can group terms to apply the theorem.

Let $y = \frac{x}{2} - \frac{2}{x}$. Then the expression is $(1 + y)^4$.

Expand $(1+y)^4$ using the Binomial Theorem with $a=1$, $b=y$, $n=4$:

$(1+y)^4 = \binom{4}{0}1^4 y^0 + \binom{4}{1}1^3 y^1 + \binom{4}{2}1^2 y^2 + \binom{4}{3}1^1 y^3 + \binom{4}{4}1^0 y^4$

$(1+y)^4 = 1 + 4y + 6y^2 + 4y^3 + y^4$

---

Now substitute back $y = \frac{x}{2} - \frac{2}{x}$ into this expansion.

$y = \frac{x}{2} - \frac{2}{x}$

$y^2 = \left(\frac{x}{2} - \frac{2}{x}\right)^2 = \left(\frac{x}{2}\right)^2 - 2\left(\frac{x}{2}\right)\left(\frac{2}{x}\right) + \left(\frac{2}{x}\right)^2 = \frac{x^2}{4} - 2 + \frac{4}{x^2}$

$y^3 = y \cdot y^2 = \left(\frac{x}{2} - \frac{2}{x}\right) \left(\frac{x^2}{4} - 2 + \frac{4}{x^2}\right)$

$y^3 = \frac{x}{2}\left(\frac{x^2}{4} - 2 + \frac{4}{x^2}\right) - \frac{2}{x}\left(\frac{x^2}{4} - 2 + \frac{4}{x^2}\right)$

$y^3 = \left(\frac{x^3}{8} - x + \frac{2}{x}\right) - \left(\frac{x}{2} - \frac{4}{x} + \frac{8}{x^3}\right)$

$y^3 = \frac{x^3}{8} - x + \frac{2}{x} - \frac{x}{2} + \frac{4}{x} - \frac{8}{x^3}$

$y^3 = \frac{x^3}{8} + (-x - \frac{x}{2}) + (\frac{2}{x} + \frac{4}{x}) - \frac{8}{x^3}$

$y^3 = \frac{x^3}{8} - \frac{3x}{2} + \frac{6}{x} - \frac{8}{x^3}$

$y^4 = (y^2)^2 = \left(\frac{x^2}{4} - 2 + \frac{4}{x^2}\right)^2$. Let $A = \frac{x^2}{4}$ and $B = \frac{4}{x^2}$. Then $y^2 = A + B - 2$.

$y^4 = ((A+B)-2)^2 = (A+B)^2 - 2(A+B)(2) + 4$

$y^4 = (A^2 + 2AB + B^2) - 4(A+B) + 4$

$y^4 = \left(\frac{x^2}{4}\right)^2 + 2\left(\frac{x^2}{4}\right)\left(\frac{4}{x^2}\right) + \left(\frac{4}{x^2}\right)^2 - 4\left(\frac{x^2}{4} + \frac{4}{x^2}\right) + 4$

$y^4 = \frac{x^4}{16} + 2(1) + \frac{16}{x^4} - x^2 - \frac{16}{x^2} + 4$

$y^4 = \frac{x^4}{16} + 2 + \frac{16}{x^4} - x^2 - \frac{16}{x^2} + 4$

$y^4 = \frac{x^4}{16} - x^2 - \frac{16}{x^2} + \frac{16}{x^4} + 6$

---

Now substitute $y$, $y^2$, $y^3$, and $y^4$ back into the expansion $1 + 4y + 6y^2 + 4y^3 + y^4$:

$1 + 4\left(\frac{x}{2} - \frac{2}{x}\right) + 6\left(\frac{x^2}{4} - 2 + \frac{4}{x^2}\right) + 4\left(\frac{x^3}{8} - \frac{3x}{2} + \frac{6}{x} - \frac{8}{x^3}\right) + \left(\frac{x^4}{16} - x^2 - \frac{16}{x^2} + \frac{16}{x^4} + 6\right)$

---

Distribute the coefficients:

$1 + \left(2x - \frac{8}{x}\right) + \left(\frac{6x^2}{4} - 12 + \frac{24}{x^2}\right) + \left(\frac{4x^3}{8} - \frac{12x}{2} + \frac{24}{x} - \frac{32}{x^3}\right) + \left(\frac{x^4}{16} - x^2 - \frac{16}{x^2} + \frac{16}{x^4} + 6\right)$

$1 + 2x - \frac{8}{x} + \frac{3x^2}{2} - 12 + \frac{24}{x^2} + \frac{x^3}{2} - 6x + \frac{24}{x} - \frac{32}{x^3} + \frac{x^4}{16} - x^2 - \frac{16}{x^2} + \frac{16}{x^4} + 6$

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Group terms by powers of $x$:

$x^4$: $\frac{x^4}{16}$

$x^3$: $\frac{x^3}{2}$

$x^2$: $\frac{3x^2}{2} - x^2 = \frac{3}{2}x^2 - \frac{2}{2}x^2 = \frac{1}{2}x^2$

$x^1$: $2x - 6x = -4x$

$x^0$ (constant): $1 - 12 + 6 = -5$

$x^{-1}$: $-\frac{8}{x} + \frac{24}{x} = \frac{-8+24}{x} = \frac{16}{x}$

$x^{-2}$: $\frac{24}{x^2} - \frac{16}{x^2} = \frac{24-16}{x^2} = \frac{8}{x^2}$

$x^{-3}$: $-\frac{32}{x^3}$

$x^{-4}$: $\frac{16}{x^4}$

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Combine all the terms:

$\frac{x^4}{16} + \frac{x^3}{2} + \frac{x^2}{2} - 4x - 5 + \frac{16}{x} + \frac{8}{x^2} - \frac{32}{x^3} + \frac{16}{x^4}$

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The expanded form is:

$\mathbf{\frac{x^4}{16} + \frac{x^3}{2} + \frac{x^2}{2} - 4x - 5 + \frac{16}{x} + \frac{8}{x^2} - \frac{32}{x^3} + \frac{16}{x^4}}$

Given expression:

$(3x^2 - 2ax + 3a^2 )^3$

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To Expand:

Expand the given trinomial using the Binomial Theorem.

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Solution:

The Binomial Theorem is used for expanding expressions with two terms. To expand a trinomial like this, we can group terms together to form a binomial.

Let's group the terms as $((3x^2) + (-2ax + 3a^2))^3$.

Let $u = 3x^2$ and $v = -2ax + 3a^2$. The expression becomes $(u + v)^3$.

Using the Binomial Theorem for $n=3$:

$(u+v)^3 = \binom{3}{0}u^3v^0 + \binom{3}{1}u^2v^1 + \binom{3}{2}u^1v^2 + \binom{3}{3}u^0v^3$

$(u+v)^3 = 1 \cdot u^3 + 3 \cdot u^2 v + 3 \cdot u v^2 + 1 \cdot v^3$

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Now substitute back $u = 3x^2$ and $v = -2ax + 3a^2$ and expand each term:

Term 1: $u^3$

$u^3 = (3x^2)^3 = 3^3 (x^2)^3 = 27x^{2 \times 3} = 27x^6$

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Term 2: $3u^2v$

$3u^2v = 3(3x^2)^2 (-2ax + 3a^2)$

$3u^2v = 3(9x^4) (-2ax + 3a^2)$

$3u^2v = 27x^4 (-2ax + 3a^2)$

$3u^2v = 27x^4 (-2ax) + 27x^4 (3a^2)$

$3u^2v = -54ax^5 + 81a^2x^4$

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Term 3: $3uv^2$

$3uv^2 = 3(3x^2) (-2ax + 3a^2)^2$

$3uv^2 = 9x^2 (-2ax + 3a^2)^2$

Expand $(-2ax + 3a^2)^2$ using $(p+q)^2 = p^2 + 2pq + q^2$ with $p=-2ax$ and $q=3a^2$:

$(-2ax + 3a^2)^2 = (-2ax)^2 + 2(-2ax)(3a^2) + (3a^2)^2$

$(-2ax + 3a^2)^2 = 4a^2x^2 - 12a^3x + 9a^4$

Substitute this back into the expression for Term 3:

$3uv^2 = 9x^2 (4a^2x^2 - 12a^3x + 9a^4)$

$3uv^2 = 9x^2 (4a^2x^2) + 9x^2 (-12a^3x) + 9x^2 (9a^4)$

$3uv^2 = 36a^2x^4 - 108a^3x^3 + 81a^4x^2$

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Term 4: $v^3$

$v^3 = (-2ax + 3a^2)^3$

Expand $(-2ax + 3a^2)^3$ using $(p+q)^3 = p^3 + 3p^2q + 3pq^2 + q^3$ with $p=-2ax$ and $q=3a^2$:

$v^3 = (-2ax)^3 + 3(-2ax)^2(3a^2) + 3(-2ax)(3a^2)^2 + (3a^2)^3$

$v^3 = -8a^3x^3 + 3(4a^2x^2)(3a^2) + 3(-2ax)(9a^4) + 27a^6$

$v^3 = -8a^3x^3 + 36a^4x^2 - 54a^5x + 27a^6$

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Add all the expanded terms together:

$(3x^2 - 2ax + 3a^2)^3 = (27x^6) + (-54ax^5 + 81a^2x^4) + (36a^2x^4 - 108a^3x^3 + 81a^4x^2) + (-8a^3x^3 + 36a^4x^2 - 54a^5x + 27a^6)$

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Combine like terms (terms with the same powers of $x$ and $a$). Order by descending powers of $x$:

$27x^6$

$-54ax^5$

$81a^2x^4 + 36a^2x^4 = (81+36)a^2x^4 = 117a^2x^4$

$-108a^3x^3 - 8a^3x^3 = (-108-8)a^3x^3 = -116a^3x^3$

$81a^4x^2 + 36a^4x^2 = (81+36)a^4x^2 = 117a^4x^2$

$-54a^5x$

$27a^6$

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The final expanded form is:

$\mathbf{27x^6 - 54ax^5 + 117a^2x^4 - 116a^3x^3 + 117a^4x^2 - 54a^5x + 27a^6}$