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| Development of Binomial Theorem | Binomial Theorem for Positive Integral Index | General and Middle Terms |
Chapter 8 Binomial Theorem (Concepts)
This chapter introduces a powerful algebraic tool known as the Binomial Theorem, providing a systematic and efficient formula for expanding binomial expressions raised to any positive integer power. Recall that expanding expressions like $(a+b)^2$ or $(a+b)^3$ is manageable through direct multiplication ($a^2+2ab+b^2$ and $a^3+3a^2b+3ab^2+b^3$, respectively). However, attempting to expand $(a+b)^n$ for larger integer values of $n$, say $(a+b)^7$ or $(a+b)^{10}$, using repeated multiplication becomes exceedingly tedious and prone to errors. The Binomial Theorem offers an elegant and direct formula to achieve this expansion systematically for any positive integer $n$.
The theorem states that the expansion of $(a+b)^n$ is given by a sum of terms, where each term involves powers of $a$ and $b$ multiplied by specific coefficients. The formula is: $$ (a + b)^n = {^{n}C_{0}} a^n b^0 + {^{n}C_{1}} a^{n-1} b^1 + {^{n}C_{2}} a^{n-2} b^2 + \ $$ $$ \dots + {^{n}C_{r}} a^{n-r} b^r + \dots + {^{n}C_{n}} a^0 b^n $$ Using concise summation notation, this can be written as: $$ \mathbf{(a + b)^n = \sum\limits_{r=0}^{n} {^{n}C_{r}} a^{n-r} b^r} $$ The coefficients $\mathbf{{^{n}C_{r}}}$ (read as "n choose r") are known as the Binomial Coefficients. These are precisely the same values encountered in combinatorics when calculating combinations, given by the formula: $$ \mathbf{{^{n}C_{r}} = \binom{n}{r} = \frac{n!}{r!(n - r)!}} $$ where $n!$ denotes the factorial of $n$.
Several key patterns and observations emerge from this expansion, which are crucial for understanding and application:
- The total number of terms in the expansion of $(a+b)^n$ is always $\mathbf{n+1}$.
- The powers of the first term, $a$, start at $n$ in the first term and decrease by 1 in each subsequent term, down to $0$ in the last term ($a^n, a^{n-1}, \dots, a^1, a^0$).
- Conversely, the powers of the second term, $b$, start at $0$ in the first term and increase by 1 in each subsequent term, up to $n$ in the last term ($b^0, b^1, \dots, b^{n-1}, b^n$).
- In every term of the expansion, the sum of the powers of $a$ and $b$ is always equal to $n$. (e.g., in the term ${^nC_r} a^{n-r} b^r$, the sum of powers is $(n-r) + r = n$).
- The binomial coefficients are symmetric: the coefficient of the $r^{th}$ term from the beginning is equal to the coefficient of the $r^{th}$ term from the end, because ${^nC_r} = {^nC_{n-r}}$.
A visually appealing way to generate binomial coefficients for smaller values of $n$ is Pascal's Triangle. This triangular array starts with 1 at the top. Each subsequent row starts and ends with 1, and every other entry is obtained by summing the two entries directly above it in the preceding row. The $(n+1)^{th}$ row of Pascal's Triangle provides the coefficients ${^nC_0, ^nC_1, \dots, ^nC_n}$ for the expansion of $(a+b)^n$.
Beyond the full expansion, several important applications arise:
- Finding the General Term: The term involving $b^r$ is the $(r+1)^{th}$ term in the expansion (since $r$ starts from 0). This term, denoted $T_{r+1}$, is given by the formula: $\mathbf{T_{r+1} = {^{n}C_{r}} a^{n-r} b^r}$. This is extremely useful for finding a specific term (e.g., the 5th term, where $r=4$), the term independent of a variable (constant term), or the coefficient of a specific power (e.g., coefficient of $x^k$).
- Determining the Middle Term(s):
- If the exponent $n$ is even, there is exactly one middle term, which is the $\left(\frac{n}{2} + 1\right)^{th}$ term (found by setting $r = \frac{n}{2}$).
- If the exponent $n$ is odd, there are two middle terms: the $\left(\frac{n+1}{2}\right)^{th}$ term (using $r = \frac{n-1}{2}$) and the $\left(\frac{n+1}{2} + 1\right)^{th}$ term (using $r = \frac{n+1}{2}$).
- The theorem is easily adapted for expansions like $(a - b)^n$ by substituting $(-b)$ for $b$, resulting in alternating signs in the expansion. It can also be used for numerical approximations, for instance, calculating $(1.01)^5$ by expanding $(1 + 0.01)^5$.
The Binomial Theorem is thus a fundamental tool for algebraic expansion and analysis.
Development of Binomial Theorem
The Binomial Theorem is a powerful algebraic tool used to expand expressions of the form $(a + b)^n$, where $n$ is a positive integer. In elementary algebra, we are familiar with the squares and cubes of binomials such as $(a + b)^2$ and $(a + b)^3$. These identities allow us to calculate values like $103^2$ by expressing it as $(100 + 3)^2$.
However, as the index (power) increases, such as in $(103)^7$ or $(998)^9$, the process of repeated multiplication becomes extremely tedious and prone to errors. To overcome this difficulty, the general result known as the Binomial Theorem was developed. This theorem, first introduced by Sir Isaac Newton, provides a systematic way to expand any power of a binomial expression.
Expansions for Positive Integral Indices
The expansion of $(a + b)^n$ is obtained by repeated algebraic multiplication. Let us examine the step-by-step derivation for the initial indices:
Derivation of Base Cases
For $n=0$:
By the laws of exponents, any non-zero base raised to the power of zero is unity.
$(a + b)^0 = 1$
[Power Rule: $x^0 = 1$]
For $n=1$:
$(a + b)^1 = a + b$
For $n=2$:
Multiplying $(a+b)$ by itself:
$(a + b)^2 = (a+b)(a+b)$
$= a(a+b) + b(a+b)$
[Distributive Law]
$= a^2 + ab + ba + b^2$
$= a^2 + 2ab + b^2$
[Since $ab = ba$]
For $n=3$:
Multiplying the result of $(a+b)^2$ by $(a+b)$:
$(a + b)^3 = (a^2 + 2ab + b^2)(a + b)$
$= a(a^2 + 2ab + b^2) + b(a^2 + 2ab + b^2)$
$= a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3$
$= a^3 + 3a^2b + 3ab^2 + b^3$
Comprehensive Analysis of Patterns
The expansions reveal three fundamental properties that hold true for any positive integral index $n$. These properties are essential for solving complex algebraic problems in competitive exams.
| Index ($n$) | Expansion | Number of Terms ($n+1$) | Sum of Powers |
|---|---|---|---|
| 1 | $a + b$ | 2 | 1 |
| 2 | $a^2 + 2ab + b^2$ | 3 | 2 |
| 3 | $a^3 + 3a^2b + 3ab^2 + b^3$ | 4 | 3 |
| $n$ | ${}^nC_0 a^n + \dots + {}^nC_n b^n$ | $n+1$ | $n$ |
1. The Rule of Term Count
In the expansion of $(a + b)^n$, the total number of terms is exactly one more than the index. This is because the expansion starts with the power of $b$ as 0 and ends with the power of $b$ as $n$, leading to $n+1$ distinct terms.
2. The Law of Decreasing and Increasing Powers
Consider the expansion of $(a + b)^n$:
• The index of the first term '$a$' starts at $n$ and decreases by 1 in each subsequent term: $n, (n-1), (n-2), \dots, 1, 0$.
• The index of the second term '$b$' starts at 0 and increases by 1 in each subsequent term: $0, 1, 2, \dots, (n-1), n$.
3. The Principle of Homogeneity (Sum of Indices)
In every term of the expansion, the sum of the indices of $a$ and $b$ is constant and always equal to $n$. Mathematically, for the $(r+1)^{th}$ term, the powers are $a^{n-r}$ and $b^r$. Their sum is:
$\text{Sum} = (n - r) + r = n$
Pascal's Triangle
The Pascal’s Triangle is not merely a geometric arrangement of numbers; it serves as a visual representation of the coefficients of the terms in the expansion of a binomial. For any positive integral index $n$, the coefficients of $(a+b)^n$ are precisely the numbers found in the $n^{th}$ row of this triangle.
Relevance of Pascal's Triangle to Binomial Expansion
When we expand $(a+b)^n$, the coefficients of the terms follow a specific symmetric pattern. Pascal's Triangle allows us to determine these coefficients without manual multiplication. Let us look at how the triangle relates to the expansion:
1. For $(a+b)^0$: The coefficient is 1 (Row 0).
2. For $(a+b)^1$: The coefficients are 1 and 1 (Row 1).
3. For $(a+b)^2$: The coefficients are 1, 2, and 1 (Row 2).
4. For $(a+b)^3$: The coefficients are 1, 3, 3, and 1 (Row 3).
Algebraic Connection: The Addition Property
The most significant link between the Binomial Theorem and Pascal's Triangle is the Addition Property. In the triangle, each number is the sum of the two numbers directly above it. In algebraic terms, this is represented by Pascal's Rule for combinations:
${}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r$
[Pascal's Identity]
This identity explains why the $(r+1)^{th}$ coefficient in the $(n+1)^{th}$ row is the sum of the $r^{th}$ and $(r+1)^{th}$ coefficients of the $n^{th}$ row.
Visualizing the Coefficients
The following table illustrates how the combination values ${}^nC_r$ map directly onto the entries of Pascal's Triangle:
| Index ($n$) | Combinatorial Form of Coefficients |
|---|---|
| 0 | ${}^0C_0$ |
| 1 | ${}^1C_0 \quad {}^1C_1$ |
| 2 | ${}^2C_0 \quad {}^2C_1 \quad {}^2C_2$ |
| 3 | ${}^3C_0 \quad {}^3C_1 \quad {}^3C_2 \quad {}^3C_3$ |
| $n$ | ${}^nC_0 \quad {}^nC_1 \quad {}^nC_2 \quad \dots \quad {}^nC_n$ |
Properties of the Triangle
1. Symmetry
The triangle is symmetric about its vertical axis. This is because of the property:
${}^nC_r = {}^nC_{n-r}$
(Property of Symmetry)
This means the first coefficient is equal to the last, the second is equal to the second last, and so on.
2. The Sum of Rows
If we add all the numbers in a specific row $n$, the sum is always a power of 2. Specifically, the sum of all binomial coefficients in the $n^{th}$ row is $2^n$.
$\sum\limits_{r=0}^{n} {}^nC_r = 2^n$
[Row Sum Formula]
Example 1. Using Pascal's Triangle, determine the expansion of $(x + y)^4$.
Answer:
Step 1: Identify the coefficients for $n=4$.
Looking at the 4th row of Pascal's Triangle (after row 0), the coefficients are: 1, 4, 6, 4, 1.
Step 2: Assign powers to $x$ and $y$.
The powers of $x$ decrease from 4 to 0, and the powers of $y$ increase from 0 to 4.
Step 3: Combine coefficients with variables.
$(x + y)^4 = 1(x^4 y^0) + 4(x^3 y^1) + 6(x^2 y^2) + 4(x^1 y^3) + 1(x^0 y^4)$
Step 4: Simplify.
$(x + y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$
Example 2. Prove that the sum of the coefficients in the expansion of $(a+b)^3$ is 8.
Answer:
To find the sum of coefficients in any binomial expansion, we can substitute the variables with 1.
Let $a = 1$ and $b = 1$.
$(1 + 1)^3 = 2^3 = 8$
Alternatively (using Pascal's Triangle):
The coefficients for $n=3$ are 1, 3, 3, and 1.
$\text{Sum} = 1 + 3 + 3 + 1 = 8$
Since $8 = 2^3$, the property $\sum {}^nC_r = 2^n$ is verified.
Binomial Coefficients using Combinations
While Pascal's Triangle is an excellent visual aid for smaller powers, it becomes impractical for large indices such as $(x+y)^{50}$. To calculate coefficients for any $n$, we utilize the Combinatorial Method. This approach relies on the concept of selections, denoted by the symbol ${}^nC_r$.
Factorial Notation
Before defining binomial coefficients, it is essential to understand Factorials. The factorial of a non-negative integer $n$, denoted by $n!$, is the product of all positive integers less than or equal to $n$.
$n! = n \times (n-1) \times (n-2) \times \dots \times 3 \times 2 \times 1$
Table of Factorial Values
| Number ($n$) | Factorial ($n!$) |
|---|---|
| 0 | 1 |
| 1 | 1 |
| 2 | 2 |
| 3 | 6 |
| 4 | 24 |
| 5 | 120 |
Note: By definition, $0! = 1$. This ensures that the combinatorial formulas remain consistent.
The Combination Formula
The binomial coefficient ${}^nC_r$ (also written as $\binom{n}{r}$) represents the number of ways to select $r$ items from a set of $n$ distinct items. It is mathematically defined as:
${}^nC_r = \frac{n!}{r!(n-r)!}$
[where $0 \leq r \leq n$]
Derivation of Property: Symmetry
The symmetry observed in Pascal's triangle (where the first coefficient equals the last) can be proven using the formula:
To Prove: ${}^nC_r = {}^nC_{n-r}$
Proof:
From the general formula:
L.H.S = ${}^nC_r = \frac{n!}{r!(n-r)!}$
Now, calculating for $(n-r)$:
R.H.S = ${}^nC_{n-r} = \frac{n!}{(n-r)!(n-(n-r))!}$
$= \frac{n!}{(n-r)! r!}$
Since the denominators are identical, L.H.S = R.H.S. Hence, the coefficients are symmetric.
Algebraic Derivation of Pascal's Rule
Pascal's Rule states that the sum of two adjacent coefficients in one row equals a coefficient in the next row. This is the logic behind the construction of the triangle.
${}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r$
Proof:
L.H.S = $\frac{n!}{r!(n-r)!} + \frac{n!}{(r-1)!(n-r+1)!}$
Taking out the common terms from the factorials:
$= \frac{n!}{(r-1)!(n-r)!} \left[ \frac{1}{r} + \frac{1}{n-r+1} \right]$
Finding the common denominator inside the bracket:
$= \frac{n!}{(r-1)!(n-r)!} \left[ \frac{n-r+1+r}{r(n-r+1)} \right]$
$= \frac{n! \times (n+1)}{(r-1)! \cdot r \times (n-r)! \cdot (n-r+1)}$
Using the property $k \times (k-1)! = k!$:
$= \frac{(n+1)!}{r!(n+1-r)!} = {}^{n+1}C_r$
[Hence Proved]
Example 1. Calculate the value of the binomial coefficient ${}^7C_3$.
Answer:
Using the formula ${}^nC_r = \frac{n!}{r!(n-r)!}$ where $n=7$ and $r=3$:
${}^7C_3 = \frac{7!}{3!(7-3)!} = \frac{7!}{3! \times 4!}$
Expanding the factorials to cancel common terms:
${}^7C_3 = \frac{7 \times 6 \times 5 \times \cancel{4!}}{(3 \times 2 \times 1) \times \cancel{4!}}$
${}^7C_3 = \frac{7 \times \cancel{6} \times 5}{\cancel{6}} = 7 \times 5 = 35$
Final Answer: The value of ${}^7C_3$ is 35.
Example 2. If ${}^nC_8 = {}^nC_{12}$, find the value of ${}^nC_{24}$.
Answer:
Given: ${}^nC_8 = {}^nC_{12}$
We know the property: If ${}^nC_x = {}^nC_y$, then either $x=y$ or $x+y=n$.
Since $8 \neq 12$, we must have:
$n = 8 + 12$
$n = 20$
Now, we need to find ${}^nC_{24}$, which is ${}^{20}C_{24}$.
However, the condition for combination is $0 \leq r \leq n$. Since $24 > 20$, the selection is not possible.
Result: In standard real-valued combinatorics, ${}^{20}C_{24} = 0$.
Example 3. A student needs to select 2 out of 5 subjects for an elective. How many combinations are possible? Represent this as a binomial coefficient.
Answer:
The number of ways to select 2 subjects out of 5 is given by ${}^5C_2$.
${}^5C_2 = \frac{5!}{2!(5-2)!} = \frac{5!}{2! \times 3!}$
$= \frac{5 \times 4 \times \cancel{3!}}{2 \times 1 \times \cancel{3!}}$
$= \frac{20}{2} = 10$
There are 10 possible combinations.
The General Binomial Theorem
The General Binomial Theorem provides a consolidated formula to expand $(a+b)^n$ for any positive integer $n$. By replacing the manual coefficients derived from Pascal's triangle with their corresponding combinatorial values, we achieve a universal algebraic expansion. This theorem is foundational in higher mathematics, probability, and calculus.
Statement of the Theorem
For any positive integer $n$ and any real numbers $a$ and $b$, the expansion is given by:
$(a+b)^n = {}^nC_0 a^n b^0 + {}^nC_1 a^{n-1}b^1 + {}^nC_2 a^{n-2}b^2 \ $$ + \dots + {}^nC_r a^{n-r}b^r + \dots + {}^nC_n a^0 b^n$
Using Sigma ($\sum$) notation, the expansion is expressed compactly as:
$(a+b)^n = \sum\limits_{r=0}^{n} {}^nC_r a^{n-r} b^r$
Note: This theorem is formally proven using the Principle of Mathematical Induction (PMI).
General Term of the Expansion
Instead of expanding the entire binomial expression, it is often required to find a single specific term. The $(r+1)^{th}$ term in the expansion is denoted by $T_{r+1}$.
$T_{r+1} = {}^nC_r a^{n-r} b^r$
[General Term Formula]
We use $r+1$ instead of $r$ because the expansion starts with $r=0$ for the 1st term ($T_1$). Therefore, for the $k^{th}$ term, the value of $r$ will be $k-1$.
Middle Term(s) of the Expansion
The total number of terms in the expansion of $(a+b)^n$ is $n+1$. The position of the middle term depends entirely on whether $n$ is even or odd.
Case 1: When $n$ is Even
If $n$ is even, the total number of terms $(n+1)$ is odd. Therefore, there is only one middle term.
$\text{Middle Term} = T_{\left(\frac{n}{2} + 1\right)}$
Case 2: When $n$ is Odd
If $n$ is odd, the total number of terms $(n+1)$ is even. In this case, there are two middle terms.
$\text{Middle Terms} = T_{\left(\frac{n+1}{2}\right)} \text{ and } T_{\left(\frac{n+3}{2}\right)}$
Term Independent of Variables (Constant Term)
In certain problems, you are asked to find the "Term Independent of $x$" (or the Constant Term). This refers to the term where the power of the variable becomes zero, leaving only the coefficients.
Steps to find the independent term:
1. Write the general term $T_{r+1}$ for the given expansion.
2. Combine all the powers of the variable (say $x$).
3. Equate the resultant power of $x$ to 0.
4. Solve for $r$ and substitute it back into the general term formula.
Example 1. Find the $4^{th}$ term in the expansion of $(x - 2y)^{12}$.
Answer:
Given: $a = x$, $b = -2y$, $n = 12$. We need to find $T_4$.
To find the $4^{th}$ term, we set $r+1 = 4$, which gives $r = 3$.
Using the General Term Formula:
$T_4 = {}^{12}C_3 (x)^{12-3} (-2y)^3$
Calculating ${}^{12}C_3$:
${}^{12}C_3 = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$
Now, substituting the values:
$T_4 = 220 \times x^9 \times (-8y^3)$
$T_4 = -1760 x^9 y^3$
Final Answer: The $4^{th}$ term is $-1760 x^9 y^3$.
Example 2. Find the middle term in the expansion of $\left(\frac{x}{3} + 9y\right)^{10}$.
Answer:
Given: $n = 10$, which is an even number. Therefore, there is only one middle term.
$\text{Middle Term} = T_{\left(\frac{10}{2} + 1\right)} = T_6$
For $T_6$, we take $r = 5$.
$a = \frac{x}{3}$ and $b = 9y$.
$T_6 = {}^{10}C_5 \left(\frac{x}{3}\right)^{10-5} (9y)^5$
$T_6 = {}^{10}C_5 \left(\frac{x}{3}\right)^5 (9y)^5$
Using the shortcut product calculation for ${}^{10}C_5$:
${}^{10}C_5 = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$
Now simplify the variables:
$T_6 = 252 \times \frac{x^5}{3^5} \times 9^5 y^5$
Since $9 = 3^2$, $9^5 = (3^2)^5 = 3^{10}$.
$T_6 = 252 \times x^5 \times \frac{3^{10}}{3^5} \times y^5$
$T_6 = 252 \times 3^5 \times x^5 y^5$
$T_6 = 252 \times 243 \times x^5 y^5$
$T_6 = 61236 x^5 y^5$
Example 3. Find the term independent of $x$ in the expansion of $\left(x^2 - \frac{1}{x}\right)^9$.
Answer:
Given: $a = x^2$, $b = -\frac{1}{x}$, $n = 9$.
Let the $(r+1)^{th}$ term be independent of $x$.
$T_{r+1} = {}^9C_r (x^2)^{9-r} \left(-\frac{1}{x}\right)^r$
Separate the constants and the variables:
$T_{r+1} = {}^9C_r \times (-1)^r \times (x)^{18-2r} \times (x)^{-r}$
$T_{r+1} = {}^9C_r \times (-1)^r \times x^{18-3r}$
For the term to be independent of $x$, the power of $x$ must be zero.
$18 - 3r = 0$
$3r = 18 \implies r = 6$
So, the $7^{th}$ term ($T_7$) is independent of $x$.
Substitute $r=6$ in the general term:
$T_7 = {}^9C_6 \times (-1)^6 \times x^0$
$T_7 = {}^9C_6 \times 1 \times 1$
Using symmetry, ${}^9C_6 = {}^9C_3$:
${}^9C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$
Final Answer: The term independent of $x$ is 84.
Binomial Theorem for Positive Integral Index
The Binomial Theorem provides a general formula for the expansion of any power of a binomial expression. For any natural number $n$ and any real numbers $a$ and $b$, the expansion of $(a + b)^n$ is a sum of $(n + 1)$ terms, where each term consists of a binomial coefficient and powers of $a$ and $b$.
The Mathematical Statement
If $n \in N$, then the expansion of $(a + b)^n$ is:
$(a + b)^n = {}^nC_0 a^n b^0 + {}^nC_1 a^{n-1}b^1 + {}^nC_2 a^{n-2}b^2 $$ + \dots + {}^nC_r a^{n-r}b^r + \dots + {}^nC_n a^0 b^n$
Using the compact Sigma ($\sum$) notation, it can be expressed as:
$(a + b)^n = \sum\limits_{r=0}^{n} {}^nC_r a^{n-r}b^r$
Where ${}^nC_r = \frac{n!}{r!(n-r)!}$ are called Binomial Coefficients.
Proof of the Theorem
We prove this theorem using the Principle of Mathematical Induction (PMI).
Step 1: To prove for $n = 1$
L.H.S. $= (a + b)^1 = a + b$
R.H.S. $= {}^1C_0 a^1 b^0 + {}^1C_1 a^0 b^1 = 1 \cdot a + 1 \cdot b = a + b$
L.H.S. = R.H.S.
(Hence true for $n=1$)
Step 2: To assume for $n = m$
Let the theorem be true for $n = m$, where $m$ is a positive integer.
$(a + b)^m = {}^mC_0 a^m + {}^mC_1 a^{m-1}b + {}^mC_2 a^{m-2}b^2 + \dots + {}^mC_r a^{m-r}b^r $$ + \dots + {}^mC_m b^m$
Step 3: To prove for $n = m + 1$
Consider $(a + b)^{m+1} = (a + b) \cdot (a + b)^m$. By substituting the value from the above equation:
$(a + b)^{m+1} = (a + b) [ {}^mC_0 a^m + {}^mC_1 a^{m-1}b + {}^mC_2 a^{m-2}b^2 + \dots + {}^mC_m b^m ]$
Multiplying the whole expansion by $a$ first, then by $b$:
$ = [ {}^mC_0 a^{m+1} + {}^mC_1 a^{m}b + {}^mC_2 a^{m-1}b^2 + \dots + {}^mC_m a b^m ]$
$ + [ {}^mC_0 a^m b + {}^mC_1 a^{m-1}b^2 + {}^mC_2 a^{m-2}b^3 + \dots + {}^mC_m b^{m+1} ]$
Grouping terms with identical powers of $a$ and $b$:
$ = {}^mC_0 a^{m+1} + ({}^mC_1 + {}^mC_0) a^m b + ({}^mC_2 + {}^mC_1) a^{m-1} b^2 + \dots + {}^mC_m b^{m+1}$
Using Pascal's Identity: ${}^mC_r + {}^mC_{r-1} = {}^{m+1}C_r$
${}^mC_0 = {}^{m+1}C_0 = 1$
${}^mC_m = {}^{m+1}C_{m+1} = 1$
The final expansion becomes:
$(a + b)^{m+1} = {}^{m+1}C_0 a^{m+1} + {}^{m+1}C_1 a^m b + {}^{m+1}C_2 a^{m-1} b^2 $$ + \dots + {}^{m+1}C_{m+1} b^{m+1}$
This shows that $P(m+1)$ is true. Hence, by PMI, the theorem is true for all $n \in N$.
Important Observations
1. Number of Terms
The total number of terms in the expansion of $(a + b)^n$ is $(n + 1)$. This is one more than the index of the binomial.
2. Sum of Indices
In every term of the expansion, the sum of the indices (powers) of $a$ and $b$ is constant and equal to $n$.
$\text{Power of } a + \text{Power of } b = (n-r) + r = n$
3. Symmetry of Coefficients
The coefficients of terms equidistant from the beginning and end are equal because of the property:
${}^nC_r = {}^nC_{n-r}$
The General and Middle Terms
The General Term in the expansion is the $(r+1)^{th}$ term, denoted by $T_{r+1}$:
$T_{r+1} = {}^nC_r a^{n-r} b^r$
The Middle Term depends on the value of $n$:
• If $n$ is Even, there is only one middle term: $T_{\left(\frac{n}{2} + 1\right)}$.
• If $n$ is Odd, there are two middle terms: $T_{\left(\frac{n+1}{2}\right)}$ and $T_{\left(\frac{n+3}{2}\right)}$.
Example 1. Expand $(2x - 3y)^4$.
Answer:
Using the formula $(a+b)^n = \sum\limits_{r=0}^{n} {}^nC_r a^{n-r}b^r$ where $a = 2x$, $b = -3y$, and $n=4$.
Step 1: Identify terms
The expansion will have $4+1=5$ terms.
$(2x - 3y)^4 = {}^4C_0 (2x)^4 + {}^4C_1 (2x)^3(-3y) + {}^4C_2 (2x)^2(-3y)^2 $$ + {}^4C_3 (2x)^1(-3y)^3 + {}^4C_4 (-3y)^4$
Step 2: Calculate Binomial Coefficients and Powers
${}^4C_0 = 1, {}^4C_1 = 4, {}^4C_2 = 6, {}^4C_3 = 4, {}^4C_4 = 1$
Step 3: Substitute
$(2x - 3y)^4 = 1(16x^4) + 4(8x^3)(-3y) + 6(4x^2)(9y^2) $$ + 4(2x)(-27y^3) + 1(81y^4)$
$(2x - 3y)^4 = 16x^4 - 96x^3y + 216x^2y^2 - 216xy^3 + 81y^4$
Example 2. A shopkeeper calculates the compound interest on a sum using the expression $\textsf{₹} 10,000 \times (1.02)^5$. Expand $(1.02)^5$ using the binomial theorem to find the multiplier.
Answer:
To find the multiplier, we can express $1.02$ as a binomial $(1 + 0.02)$. Here, $a = 1$, $b = 0.02$, and $n = 5$. Since $n=5$, there will be $5+1=6$ terms in the expansion.
Step 1: Write the full expansion using the General Binomial Theorem.
$(1 + 0.02)^5 = {}^5C_0(1)^5(0.02)^0 + {}^5C_1(1)^4(0.02)^1 + {}^5C_2(1)^3(0.02)^2 \ $$ + {}^5C_3(1)^2(0.02)^3 + {}^5C_4(1)^1(0.02)^4 + {}^5C_5(1)^0(0.02)^5$
Step 2: Calculate binomial coefficients and simplified terms.
Using the coefficients from Pascal's Triangle for $n=5$, which are 1, 5, 10, 10, 5, 1:
$(1.02)^5 = 1(1) + 5(0.02) + 10(0.0004) + 10(0.000008) + 5(0.00000016) \ $$ + 1(0.0000000032)$
Step 3: Perform the numerical addition.
$(1.02)^5 = 1 + 0.1 + 0.004 + 0.00008 + 0.0000008 + 0.0000000032$
Summing these values together:
$(1.02)^5 = 1.1040808032$
Final Multiplier Calculation:
The multiplier is exactly 1.1040808032. For practical financial calculations in $\textsf{₹}$, this is often rounded to $1.10408$.
To find the final amount:
$\text{Amount} = 10,000 \times 1.1040808032$
$\text{Amount} = \textsf{₹} 11,040.81$
(Approx.)
Special Cases of Binomial Expansion
1. Expansion of $(a - b)^n$
By replacing $b$ with $-b$ in the standard formula:
$(a - b)^n = {}^nC_0 a^n - {}^nC_1 a^{n-1}b + {}^nC_2 a^{n-2}b^2 - \dots + (-1)^n {}^nC_n b^n$
The terms possess alternating signs. The final sign depends on whether $n$ is even ($+$) or odd ($-$).
2. Expansion of $(1 + x)^n$
Substituting $a = 1$ and $b = x$:
$(1 + x)^n = {}^nC_0 + {}^nC_1 x + {}^nC_2 x^2 + \dots + {}^nC_n x^n = \sum\limits_{r=0}^{n} {}^nC_r x^r$
3. Expansion of $(1 - x)^n$
$(1 - x)^n = {}^nC_0 - {}^nC_1 x + {}^nC_2 x^2 - \dots + (-1)^n {}^nC_n x^n = \sum\limits_{r=0}^{n} (-1)^r {}^nC_r x^r$
Properties of Binomial Coefficients
For the expansion $(1 + x)^n = C_0 + C_1 x + C_2 x^2 + \dots + C_n x^n$, the following properties are essential:
(i) Sum of all Coefficients:
By setting $x = 1$:
${}^nC_0 + {}^nC_1 + {}^nC_2 + \dots + {}^nC_n = 2^n$
(ii) Alternating Sum:
By setting $x = -1$:
${}^nC_0 - {}^nC_1 + {}^nC_2 - {}^nC_3 + \dots + (-1)^n {}^nC_n = 0$
(iii) Equivalence of Even and Odd Coefficients:
From above equation, the sum of even-positioned coefficients equals the sum of odd-positioned ones:
${}^nC_0 + {}^nC_2 + {}^nC_4 + \dots = {}^nC_1 + {}^nC_3 + {}^nC_5 + \dots = 2^{n-1}$
Term Count in Compound Expressions
For the linear combinations $(a + b)^n \pm (a - b)^n$, the number of terms after simplification is:
1. If $n$ is Odd:
Both the sum and the difference expressions contain $\frac{n+1}{2}$ terms.
2. If $n$ is Even:
• $(a + b)^n + (a - b)^n$ results in $\frac{n}{2} + 1$ terms.
• $(a + b)^n - (a - b)^n$ results in $\frac{n}{2}$ terms.
Example 3. Calculate the sum of all binomial coefficients in the expansion of $(1 + x)^{10}$.
Answer:
As per the property derived in the above equation, the sum of all binomial coefficients for index $n$ is $2^n$.
Given $n = 10$:
$S = 2^{10}$
$S = 1024$
Final Answer: 1024
General and Middle Terms
In the expansion of a binomial expression, it is often necessary to find a specific term without writing out the entire expansion. The General Term provides a mathematical formula to identify any term based on its position.
The General Term
Consider the binomial expansion for a positive integral index $n$:
$(a + b)^n = {}^nC_0 a^n + {}^nC_1 a^{n-1}b + {}^nC_2 a^{n-2}b^2 + \dots + {}^nC_r a^{n-r}b^r \ $$ + \dots + {}^nC_n b^n$
By carefully analyzing the individual terms of this expansion, we observe a consistent pattern in their structure:
First term $(T_1) = {}^nC_0 a^{n-0} b^0$
[Index of $b$ is $1-1=0$]
Second term $(T_2) = {}^nC_1 a^{n-1} b^1$
[Index of $b$ is $2-1=1$]
Third term $(T_3) = {}^nC_2 a^{n-2} b^2$
[Index of $b$ is $3-1=2$]
From this pattern, it is evident that for any term at the $k^{th}$ position, the subscript of the binomial coefficient and the power of $b$ is always $(k-1)$. Therefore, if we consider the $(r+1)^{th}$ term, the index becomes $r$.
The General Term Formula
The $(r+1)^{th}$ term, denoted by $T_{r+1}$, is called the general term of the expansion $(a+b)^n$. It is given by:
$T_{r+1} = {}^nC_r a^{n-r} b^r$
Particular Cases of General Terms
Based on the variation of the binomial expression, the general term formula undergoes specific changes:
1. In the expansion of $(a - b)^n$
Since the second term is negative, the sign of the terms alternates. The general term is:
$T_{r+1} = {}^nC_r a^{n-r} (-b)^r = (-1)^r \cdot {}^nC_r a^{n-r} b^r$
2. In the expansion of $(1 + x)^n$
Here $a=1$ and $b=x$. Since any power of 1 remains 1:
$T_{r+1} = {}^nC_r (1)^{n-r} x^r = {}^nC_r x^r$
3. In the expansion of $(1 - x)^n$
$T_{r+1} = (-1)^r \cdot {}^nC_r x^r$
Term from the End of the Expansion
Sometimes, a question asks for a term starting from the last term of the expansion. There are two ways to solve this:
Method 1: Change of Binomial Order
The $r^{th}$ term from the end in the expansion of $(a + b)^n$ is exactly the same as the $r^{th}$ term from the beginning in the expansion of $(b + a)^n$.
Method 2: Positional Formula
As the total number of terms in the expansion of $(a + b)^n$ is $(n + 1)$, the $r^{th}$ term from the end has $(r - 1)$ terms after it. Thus, its position from the beginning is calculated as:
$\text{Position from beginning} = (n + 1) - r + 1 = (n - r + 2)^{th} \text{ term}$
Middle Term or Terms
The number of middle terms depends on whether the index $n$ is even or odd. Since the expansion contains $(n + 1)$ terms:
(i) If $n$ is Even
The total number of terms $(n + 1)$ is odd. In this case, there is only one middle term.
$\text{Middle Term} = \left(\frac{n}{2} + 1\right)^{th} \text{ term}$
(ii) If $n$ is Odd
The total number of terms $(n + 1)$ is even. In this case, there are two middle terms.
$\text{Middle Terms} = \left(\frac{n+1}{2}\right)^{th} \text{ and } \left(\frac{n+3}{2}\right)^{th} \text{ terms}$
| Index ($n$) | Number of Terms ($n+1$) | Number of Middle Terms | Position of Middle Term(s) |
|---|---|---|---|
| Even (e.g., 2, 4, 6) | Odd | One | $T_{\left(\frac{n}{2} + 1\right)}$ |
| Odd (e.g., 3, 5, 7) | Even | Two | $T_{\left(\frac{n+1}{2}\right)}$ and $T_{\left(\frac{n+3}{2}\right)}$ |
Example 1. Find the $4^{th}$ term from the end in the expansion of $\left(\frac{x^3}{2} - \frac{2}{x^2}\right)^9$.
Answer:
Given: $n = 9$, $r = 4$ (from end).
Using the positional formula, the term from the beginning is:
$\text{Term from start} = (n - r + 2)^{th}$
$= (9 - 4 + 2)^{th} = 7^{th} \text{ term}$
To find $T_7$, we set $r = 6$. Here $a = \frac{x^3}{2}$ and $b = -\frac{2}{x^2}$.
$T_7 = {}^9C_6 \left(\frac{x^3}{2}\right)^{9-6} \left(-\frac{2}{x^2}\right)^6$
Calculating ${}^9C_6 = {}^9C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.
$T_7 = 84 \cdot \left(\frac{x^3}{2}\right)^3 \cdot \left(\frac{2^6}{x^{12}}\right)$
$T_7 = 84 \cdot \frac{x^9}{8} \cdot \frac{64}{x^{12}} = \frac{672}{x^3}$
Final Answer: The term is $\frac{672}{x^3}$.
Example 2. Find the middle term in the expansion of $(x + 3y)^8$.
Answer:
Here $n = 8$, which is an even integer. Total number of terms = $8 + 1 = 9$ (Odd).
There is only one middle term, which is:
$\text{Middle Term} = \left(\frac{8}{2} + 1\right)^{th} = 5^{th} \text{ term}$
To find $T_5$, we take $r = 4$, $a = x$, and $b = 3y$:
$T_5 = {}^8C_4 (x)^{8-4} (3y)^4$
${}^8C_4 = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$.
$T_5 = 70 \cdot x^4 \cdot 81y^4$
$T_5 = 5670 x^4 y^4$