Chapter 8 Binomial Theorem (Concepts)

Development of Binomial Theorem

The Binomial Theorem is a fundamental theorem of algebra that describes the algebraic expansion of powers of a binomial. A binomial is simply a polynomial with two terms, for example, $(a+b)$. The theorem provides a formula that eliminates the need for tedious, repetitive multiplication when expanding expressions like $(a+b)^n$, especially for large values of the exponent $n$.

To understand how this theorem was developed, we can start by examining the expansions for small, non-negative integer values of $n$ and look for emerging patterns.

Expansions for Small Positive Integral Indices

Let's manually expand the binomial $(a+b)$ for the first few powers.

For index $n=0$:

For index $n=1$:

For index $n=2$:

For index $n=3$:

For index $n=4$:

Identifying Patterns

By carefully analyzing the expansions above, we can identify several key patterns:

1. Number of Terms: The total number of terms in the expansion of $(a+b)^n$ is always $n+1$.

   For example, in $(a+b)^3$, $n=3$, and there are $3+1=4$ terms.

2. Powers of the First Term (a): The powers of the first term, $a$, start at $n$ and decrease by 1 in each successive term, down to 0 in the last term.

   For $(a+b)^4$, the powers of $a$ are $a^4, a^3, a^2, a^1, a^0$ (where $a^0=1$).

3. Powers of the Second Term (b): The powers of the second term, $b$, start at 0 and increase by 1 in each successive term, up to $n$ in the last term.

   For $(a+b)^4$, the powers of $b$ are $b^0, b^1, b^2, b^3, b^4$ (where $b^0=1$).

4. Sum of Powers: In every term of the expansion, the sum of the powers of $a$ and $b$ is equal to the index $n$.

   For $(a+b)^4$, consider the term $6a^2b^2$. The sum of the powers is $2+2=4$, which is the index $n$. For the term $4a^3b^1$, the sum is $3+1=4$.

5. Coefficients: The coefficients of the terms (the numerical parts) follow a symmetric and predictable pattern. Let's list them out:

   Index (n)	Coefficients in the expansion of $(a+b)^n$
   0	1
   1	1,   1
   2	1,   2,   1
   3	1,   3,   3,   1
   4	1,   4,   6,   4,   1

This triangular arrangement of coefficients is known as Pascal's Triangle.

The Connection to Combinatorics

The patterns observed in the coefficients lead to a powerful mathematical tool known as Pascal's Triangle, which is directly linked to the concept of combinations, or binomial coefficients.

Pascal's Triangle

Pascal's Triangle is a triangular array of numbers that provides the coefficients for binomial expansions. Each row corresponds to the expansion of $(a+b)^n$ for a given $n$.

Construction Rules:

• Each row starts and ends with 1.
• Every other number in a row is the sum of the two numbers directly above it in the previous row.

Binomial Coefficients

The numbers we see in Pascal's Triangle are so important that they have a formal name: Binomial Coefficients. A binomial coefficient tells you the number of ways you can choose a certain number of items from a larger group, where the order of selection doesn't matter. This concept is a cornerstone of combinatorics.

Notation and Meaning

The binomial coefficient is denoted in two common ways:

• ${}^{n}C_{r}$
• $\binom{n}{r}$

Both are read as "n choose r" and represent the number of ways to choose $r$ objects from a set of $n$ distinct objects.

The Formula

The value of a binomial coefficient is calculated using the combination formula, which involves factorials:

Let's break down the formula's components:

• $n!$ (read as "n factorial") is the product of all positive integers from 1 up to $n$. For example, $4! = 4 \times 3 \times 2 \times 1 = 24$.
• By special definition, $0! = 1$.
• $n$ represents the total number of items to choose from.
• $r$ represents the number of items being chosen.

Connection to Binomial Expansion

The most important connection for the Binomial Theorem is this: The coefficients in the expansion of $(a+b)^n$ are given precisely by the sequence of binomial coefficients:

${}^{n}C_{0}, {}^{n}C_{1}, {}^{n}C_{2}, \dots, {}^{n}C_{n-1}, {}^{n}C_{n}$

Verification for n=4

Let's verify this powerful connection for the expansion of $(a+b)^4$. From Pascal's triangle, we know the coefficients are 1, 4, 6, 4, 1. Let's see if the combination formula gives us the same values.

1. First coefficient (${}^{4}C_{0}$):

This corresponds to choosing 0 items from a set of 4.

2. Second coefficient (${}^{4}C_{1}$):

This corresponds to choosing 1 item from a set of 4.

3. Third coefficient (${}^{4}C_{2}$):

This corresponds to choosing 2 items from a set of 4.

4. Fourth coefficient (${}^{4}C_{3}$):

This corresponds to choosing 3 items from a set of 4.

5. Fifth coefficient (${}^{4}C_{4}$):

This corresponds to choosing 4 items from a set of 4.

The sequence of coefficients we calculated is 1, 4, 6, 4, 1. This perfectly matches the fourth row of Pascal's Triangle and the coefficients we found from manually expanding $(a+b)^4$. This confirms the direct relationship that forms the basis of the Binomial Theorem.

The Binomial Expression (The Binomial Theorem)

This direct link between the binomial coefficients and the expansion of $(a+b)^n$ allows us to write a single, powerful formula for the entire expansion. This formula is known as the Binomial Theorem.

For any non-negative integer $n$, the expansion of $(a+b)^n$ is given by:

$(a+b)^n = {}^{n}C_{0}a^n b^0 + {}^{n}C_{1}a^{n-1}b^1 + {}^{n}C_{2}a^{n-2}b^2 + \dots + {}^{n}C_{r}a^{n-r}b^r + \dots + {}^{n}C_{n}a^0 b^n$

Using the more compact summation notation ($\Sigma$), this can be written as:

Applying this formula to our $n=4$ example:

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

Substituting the coefficient values we calculated (1, 4, 6, 4, 1):

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

This confirms that the formula perfectly generates the full expansion.

Binomial Theorem for Positive Integral Index

The patterns observed in the expansions of $(a+b)^n$ for small positive integer values of $n$, specifically the relationship between the coefficients and binomial coefficients, lead to the formal statement of the Binomial Theorem for a positive integral index.

Statement of the Binomial Theorem

For any positive integer $n$, the expansion of the binomial $(a+b)$ raised to the power of $n$ is given by the following formula:

Using summation notation ($\sum\limits$), which provides a compact way to represent the sum of a series of terms, the Binomial Theorem can be written more concisely:

Here:

• $\sum\limits_{r=0}^{n}$ indicates the sum of terms from $r=0$ to $r=n$.
• $r$ is the index of the term, starting from 0.
• ${^{n}C_{r}}$ are the binomial coefficients, calculated using the combination formula: ${^{n}C_{r}} = \frac{n!}{r!(n-r)!}$.
• $a^{n-r} b^r$ represents the variables part of each term, where the powers of $a$ decrease from $n$ to 0 as $r$ increases from 0 to $n$, and the powers of $b$ increase from 0 to $n$. The sum of the exponents in each term is $(n-r) + r = n$.

Proofs of the Binomial Theorem

The theorem can be proven in several ways. The two most common methods are by mathematical induction and by a combinatorial argument.

Proof by Mathematical Induction

We will prove the statement $P(n): (a+b)^n = \sum\limits_{r=0}^{n} {^{n}C_{r}} a^{n-r} b^r$ for all positive integers $n$.

Step 1: Base Case

We need to show the statement is true for the smallest positive integer, $n=1$.

L.H.S. = $(a+b)^1 = a+b$.

R.H.S. = $\sum\limits_{r=0}^{1} {^{1}C_{r}} a^{1-r} b^r = {^{1}C_{0}}a^{1-0}b^0 + {^{1}C_{1}}a^{1-1}b^1 = (1)a^1(1) + (1)a^0b^1 = a+b$.

Since L.H.S. = R.H.S., the statement $P(1)$ is true.

Step 2: Inductive Hypothesis

Let's assume that the statement is true for some positive integer $k$. That is, we assume $P(k)$ is true.

Step 3: Inductive Step

We now need to prove that the statement is also true for $n=k+1$. That is, we need to prove that:

$(a+b)^{k+1} = \sum\limits_{r=0}^{k+1} {^{k+1}C_{r}} a^{k+1-r} b^r$

Starting with the L.H.S.:

Using the inductive hypothesis (iii), we substitute the expression for $(a+b)^k$:

Distributing $(a+b)$ over the summation:

Let's expand both summations to observe the terms:

First summation: ${^{k}C_{0}}a^{k+1}b^0 + {^{k}C_{1}}a^{k}b^1 + {^{k}C_{2}}a^{k-1}b^2 + \dots + {^{k}C_{k}}a^{1}b^k$

Second summation: ${^{k}C_{0}}a^{k}b^1 + {^{k}C_{1}}a^{k-1}b^2 + \dots + {^{k}C_{k-1}}a^{1}b^k + {^{k}C_{k}}a^{0}b^{k+1}$

Now, we group like terms (terms with the same powers of $a$ and $b$):

$= {^{k}C_{0}}a^{k+1}b^0 + ({^{k}C_{1}} + {^{k}C_{0}})a^kb^1 + ({^{k}C_{2}} + {^{k}C_{1}})a^{k-1}b^2 + \dots + ({^{k}C_{k}} + {^{k}C_{k-1}})a^1b^k + {^{k}C_{k}}a^0b^{k+1}$

We use Pascal's Identity, which states ${^{n}C_{r}} + {^{n}C_{r-1}} = {^{n+1}C_{r}}$. Applying this to our coefficients:

(${^{k}C_{1}} + {^{k}C_{0}}) = {^{k+1}C_{1}}$

(${^{k}C_{2}} + {^{k}C_{1}}) = {^{k+1}C_{2}}$, and so on.

Also, we know that ${^{k}C_{0}} = 1 = {^{k+1}C_{0}}$ and ${^{k}C_{k}} = 1 = {^{k+1}C_{k+1}}$.

Substituting these back into the expression:

$= {^{k+1}C_{0}}a^{k+1}b^0 + {^{k+1}C_{1}}a^kb^1 + {^{k+1}C_{2}}a^{k-1}b^2 + \dots + {^{k+1}C_{k}}a^1b^k + {^{k+1}C_{k+1}}a^0b^{k+1}$

This is precisely the binomial expansion for $(a+b)^{k+1}$, which can be written in summation notation as:

This is equal to the R.H.S. we wanted to prove. Thus, $P(k+1)$ is true whenever $P(k)$ is true.

By the principle of mathematical induction, the theorem is true for all positive integers $n$.

Combinatorial Proof

Consider the expansion of the product:

Each term in the final expansion is formed by choosing either '$a$' or '$b$' from each of the $n$ factors and multiplying them together. A general term will be of the form $k \cdot a^{n-r}b^r$, where the sum of the powers $(n-r)+r=n$. The coefficient $k$ represents the number of ways this specific term can be formed.

To get the term $a^{n-r}b^r$, we must choose '$b$' from exactly $r$ of the $n$ available factors. The remaining $(n-r)$ factors will automatically contribute an '$a$'.

The problem thus reduces to finding the number of ways to choose $r$ factors (from which we pick '$b$') out of a total of $n$ factors. This is a classic combination problem, and the number of ways is given by ${^{n}C_{r}}$.

Therefore, the coefficient of the term $a^{n-r}b^r$ is ${^{n}C_{r}}$.

Since $r$ (the number of times '$b$' is chosen) can range from 0 to $n$, the complete expansion is the sum of all such possible terms:

Key Properties of the Binomial Expansion $(a+b)^n$

Based on the Binomial Theorem formula, we can list several important properties of the expansion of $(a+b)^n$ for a positive integer $n$:

1. Number of Terms: The summation index $r$ goes from 0 to $n$, giving $n-0+1 = n+1$ values. Therefore, the expansion of $(a+b)^n$ contains exactly $n+1$ terms.
2. Sum of Exponents: In every term of the expansion, ${^{n}C_{r}} a^{n-r} b^r$, the sum of the exponents of $a$ and $b$ is $(n-r) + r = n$. This is consistent throughout the expansion.
3. Symmetry of Coefficients: The binomial coefficients ${^{n}C_{r}}$ have the property ${^{n}C_{r}} = {^{n}C_{n-r}}$. This means the coefficients of terms that are equidistant from the beginning and the end of the expansion are equal.
   • Coefficient of 1st term (${^{n}C_{0}}$) = Coefficient of last term (${^{n}C_{n}}$) = 1.
   • Coefficient of 2nd term (${^{n}C_{1}}$) = Coefficient of 2nd last term (${^{n}C_{n-1}}$) = n.
4. Sum of Coefficients: To find the sum of all coefficients in the expansion, we set $a=1$ and $b=1$ in the Binomial Theorem:

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

   $(2)^n = \sum\limits_{r=0}^{n} {^{n}C_{r}}$

   $\sum\limits_{r=0}^{n} {^{n}C_{r}} = {^{n}C_{0}} + {^{n}C_{1}} + \dots + {^{n}C_{n}} = 2^n$

   The sum of the binomial coefficients for a given $n$ is equal to $2^n$.

5. Expansion of $(a-b)^n$: This can be obtained by substituting $-b$ for $b$ in the standard formula.

   $(a-b)^n = \sum\limits_{r=0}^{n} {^{n}C_{r}} a^{n-r} (-b)^r = \sum\limits_{r=0}^{n} (-1)^r {^{n}C_{r}} a^{n-r} b^r$

   ... (iv)

   The term $(-1)^r$ causes the signs of the terms to alternate. The expansion is:

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

Answer:

Answer:

General and Middle Terms

While the Binomial Theorem provides the entire expansion of $(a+b)^n$, this can be cumbersome if we only need one specific term, especially for a large index $n$. The General Term is a powerful formula that acts as a blueprint for any term in the expansion, allowing us to calculate it directly without finding any other terms.

Understanding the Notation: $T_{r+1}$

The binomial expansion is a sum that starts with an index $r=0$. Let's look at the sequence of terms:

• The 1st term ($T_1$) is generated when $r=0$: ${^{n}C_{0}} a^n b^0$.
• The 2nd term ($T_2$) is generated when $r=1$: ${^{n}C_{1}} a^{n-1} b^1$.
• The 3rd term ($T_3$) is generated when $r=2$: ${^{n}C_{2}} a^{n-2} b^2$.

Notice that the term number is always one more than the value of $r$. This relationship is why we denote the general term as $T_{r+1}$. This convention ensures that the subscript of the term matches the summation index $r$. So, the $(r+1)^{th}$ term corresponds to the index $r$ in the binomial coefficient ${^{n}C_{r}}$.

To find a specific term, say the $k^{th}$ term, you must first find the corresponding value of $r$ by solving the equation $r+1 = k$, which gives $r = k-1$.

The Formula and its Components

The formula for the General Term in the expansion of $(a+b)^n$ is:

Let's break down each component:

• ${^{n}C_{r}}$: This is the binomial coefficient. It determines the numerical part of the term and represents the number of ways to choose $r$ instances of 'b' from the $n$ binomial factors.
• $a^{n-r}$: This is the power of the first term of the binomial. If we choose 'b' from $r$ factors, we must choose 'a' from the remaining $n-r$ factors.
• $b^r$: This is the power of the second term of the binomial. The exponent $r$ conveniently matches the lower index of the combination ${^{n}C_{r}}$.

General Term for $(a-b)^n$

To find the general term for $(a-b)^n$, we can cleverly rewrite it as $(a + (-b))^n$. Now, we apply the standard general term formula, treating '$a$' as the first term and '$-b$' as the second term.

Using the property of exponents, $(-b)^r = (-1)^r b^r$. This gives us the formula:

The factor $(-1)^r$ generates the alternating signs in the expansion. If $r$ is even, the term is positive. If $r$ is odd, the term is negative.

Answer:

Term Independent of a Variable

A term independent of a variable (also called the constant term) is a term where the net power of the variable is zero. To find it, we use the general term formula, simplify the variable's exponent in terms of $r$, and then solve for the value of $r$ that makes this exponent zero.

Answer:

The Middle Term(s)

In a binomial expansion, the middle term (or terms) are those located at the very center of the sequence of terms. The number of such terms depends on the total count of terms in the expansion, which is always $n+1$. This total count is either odd or even, depending on whether the index $n$ is even or odd.

Case 1: When the Index $n$ is Even

If the index $n$ is an even integer, the total number of terms in the expansion is $n+1$, which will be an odd number. An ordered set with an odd number of elements has exactly one element in the middle.

For example, if $n=4$, there are $4+1=5$ terms. The middle term is clearly the 3rd one.

Finding the Position: The position of this single middle term is given by the formula $\left(\frac{n+1+1}{2}\right) = \left(\frac{n+2}{2}\right) = \left(\frac{n}{2} + 1\right)^{th}$.

Finding the Term: To find this term using the general formula $T_{r+1}$, we set the term number equal to the position:

This gives us $r = \frac{n}{2}$. Substituting this value of $r$ into the general term formula $T_{r+1} = {^{n}C_{r}} a^{n-r} b^r$ yields the middle term:

Case 2: When the Index $n$ is Odd

If the index $n$ is an odd integer, the total number of terms in the expansion is $n+1$, which will be an even number. An ordered set with an even number of elements has two middle elements.

For example, if $n=5$, there are $5+1=6$ terms. The middle terms are the 3rd and 4th terms.

Finding the Positions: The positions of the two middle terms are given by $\left(\frac{n+1}{2}\right)^{th}$ and the next one, $\left(\frac{n+1}{2} + 1\right)^{th}$.

Finding the First Middle Term:

The position is $\left(\frac{n+1}{2}\right)^{th}$. To find this term, we set $r+1 = \frac{n+1}{2}$, which gives $r = \frac{n+1}{2} - 1 = \frac{n-1}{2}$. Substituting this $r$ into the general formula gives:

Finding the Second Middle Term:

The position is $\left(\frac{n+1}{2} + 1\right)^{th} = \left(\frac{n+3}{2}\right)^{th}$. To find this term, we set $r+1 = \frac{n+3}{2}$, which gives $r = \frac{n+3}{2} - 1 = \frac{n+1}{2}$. Substituting this $r$ into the general formula gives:

Therefore, when $n$ is odd, the two middle terms are $T_{\frac{n+1}{2}}$ and $T_{\frac{n+3}{2}}$.

Answer:

Answer: