Binomial Theorem

Introduction

The Binomial Theorem gives a formula for expanding $(a+b)^n$ for any positive integer index $n$.

Pascal's Triangle

The coefficients of the expansions of $(a+b)^n$ for $n=0, 1, 2, \dots$ can be arranged in a triangular form known as Pascal's Triangle.

First few rows:

• $n=0$: 1
• $n=1$: 1 1
• $n=2$: 1 2 1
• $n=3$: 1 3 3 1
• $n=4$: 1 4 6 4 1

Rule: Each number is sum of the two numbers directly above it.

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Statement of Binomial Theorem

For any positive integer $n$: $$(a+b)^n = {^nC_0} a^n + {^nC_1} a^{n-1}b + {^nC_2} a^{n-2}b^2 + \dots + {^nC_n} b^n$$

Sigma Notation: $$(a+b)^n = \sum_{k=0}^{n} {^nC_k} a^{n-k} b^k$$

Special Cases

1. $(x+1)^n$: $\sum_{r=0}^{n} {^nC_r} x^r = {^nC_0} + {^nC_1}x + {^nC_2}x^2 + \dots + {^nC_n}x^n$
2. $(1+x)^n$: ${^nC_0} + {^nC_1}x + {^nC_2}x^2 + \dots + {^nC_n}x^n$
3. $(1-x)^n$: ${^nC_0} - {^nC_1}x + {^nC_2}x^2 - \dots + (-1)^n {^nC_n}x^n$

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General Term

The term containing $a^{n-r}b^r$ is the $(r+1)^{th}$ term in the expansion of $(a+b)^n$. $$T_{r+1} = {^nC_r} a^{n-r} b^r$$

Middle Terms

The middle term depends on whether $n$ is even or odd.

Case 1: $n$ is Even

There is only one middle term: $$T_{\frac{n}{2} + 1}$$

Case 2: $n$ is Odd

There are two middle terms: $$T_{\frac{n+1}{2}} \quad \text{and} \quad T_{\frac{n+1}{2} + 1}$$

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Important properties

1. $^nC_0=^nC_n=1$
2. $^nC_r=^nC_{n-r}$
3. $^nC_r + ^nC_{r-1} = ^{n+1}C_r$
4. $^nC_r = 0$ for $r>n$
5. Sum of Coefficients: $C_0 + C_1 + C_2 + \dots + C_n = 2^n$
6. Sum of Odd/Even Coefficients: $C_0 + C_2 + C_4 + \dots = C_1 + C_3 + C_5 + \dots = 2^{n-1}$

Binomial Theorem

Example 1

Expand $(x^2 + \frac{3}{x})^4$, $x \neq 0$.

Solution: $= {^4C_0}(x^2)^4 + {^4C_1}(x^2)^3(\frac{3}{x}) + {^4C_2}(x^2)^2(\frac{3}{x})^2 + {^4C_3}(x^2)(\frac{3}{x})^3 + {^4C_4}(\frac{3}{x})^4$ $= x^8 + 12x^5 + 54x^2 + \frac{108}{x} + \frac{81}{x^4}$.

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Example 2

Find the number of terms in $(2x - 3y)^9$.

Solution: Number of terms = $n + 1 = 9 + 1 = 10$.

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Example 3

Find the 4th term in the expansion of $(x - 2y)^{12}$.

Solution: Put $r=3$ in general term $T_{r+1}$. $T_4 = {^{12}C_3} (x)^{12-3} (-2y)^3$ $= \frac{12 \times 11 \times 10}{6} x^9 (-8y^3) = -1760 x^9 y^3$.

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Example 4

Find the middle term in $(\frac{x}{3} + 9y)^{10}$.

Solution: $n=10$ (Even). Middle term is $(\frac{10}{2}+1) = 6^{th}$ term. $T_6 = {^{10}C_5} (\frac{x}{3})^5 (9y)^5 = 252 \cdot \frac{x^5}{3^5} \cdot 3^{10} y^5$ $= 252 \cdot 243 \cdot x^5 y^5 = 61236 x^5 y^5$.

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Example 5

Find the term independent of $x$ in expansion of $(x - \frac{1}{x})^{10}$.

Solution: $T_{r+1} = {^{10}C_r} x^{10-r} (-\frac{1}{x})^r = {^{10}C_r} (-1)^r x^{10-2r}$. For independence, power of $x$ must be 0. $10 - 2r = 0 \implies r = 5$. Term is $T_6 = {^{10}C_5} (-1)^5 = -252$.

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Example 6

Evaluate $(96)^3$ using Binomial Theorem.

Solution: $(100 - 4)^3 = {^3C_0}(100)^3 - {^3C_1}(100)^2(4) + {^3C_2}(100)(4^2) - {^3C_3}(4^3)$ $= 1000000 - 120000 + 4800 - 64 = 884736$.

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Example 7

Find coefficient of $x^5$ in $(x+3)^8$.

Solution: $T_{r+1} = {^8C_r} x^{8-r} 3^r$. We need $x^{8-r} = x^5 \implies 8-r=5 \implies r=3$. Coeff $= {^8C_3} \cdot 3^3 = 56 \times 27 = 1512$.

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Example 8

Show that $9^{n+1} - 8n - 9$ is divisible by 64.

Solution: $9^{n+1} = (1+8)^{n+1} = 1 + (n+1)8 + {^{n+1}C_2}8^2 + \dots$ $= 1 + 8n + 8 + 64(\dots) = 8n + 9 + 64k$. $9^{n+1} - 8n - 9 = 64k$. Divisible by 64.

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Example 9

Find the coefficient of $x^6 y^3$ in $(x+2y)^9$.

Solution: General term: ${^9C_r} x^{9-r} (2y)^r = {^9C_r} 2^r x^{9-r} y^r$. For $y^3$, put $r=3$. Coeff $= {^9C_3} \cdot 2^3 = 84 \times 8 = 672$.

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Example 10

Find the middle terms in $(3x - \frac{x^3}{6})^7$.

Solution: $n=7$ (Odd). Terms are $\frac{7+1}{2}=4^{th}$ and $5^{th}$. $T_4 = -\frac{105}{8} x^{11}$, $T_5 = \frac{35}{48} x^{13}$.

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Example 11

Using Binomial Theorem, determine which is larger: $(1.1)^{10000}$ or $1000$.

Solution: $(1+0.1)^{10000} = 1 + 10000(0.1) + \dots = 1 + 1000 + \dots > 1000$.

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Example 12

Find the sum of coefficients in $(x+y)^n$.

Solution: Put $x=1, y=1$. Sum $= (1+1)^n = 2^n$.

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Example 13

Find the coefficient of $x$ in expansion of $(1-2x^3+3x^5)(1+1/x)^8$.

Solution: Requires selecting terms from $(1+1/x)^8$ that multiply with bracket to give $x$. After calculation, Ans: 154.

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Example 14

Evaluate $(\sqrt{2}+1)^6 + (\sqrt{2}-1)^6$.

Solution: $= 2[C_0(\sqrt{2})^6 + C_2(\sqrt{2})^4 + C_4(\sqrt{2})^2 + C_6]$. $= 2[8 + 15(4) + 15(2) + 1] = 198$.

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Example 15

If the coefficients of $2^{nd}, 3^{rd}$ and $4^{th}$ terms in $(1+x)^{2n}$ are in A.P., show that $2n^2 - 9n + 7 = 0$.

Solution: Given $^{2n}C_1, ^{2n}C_2, ^{2n}C_3$ are in A.P. $2(^{2n}C_2) = ^{2n}C_1 + ^{2n}C_3$. Solving yields the quadratic equation.

Questions and Answers (Board Exam)

Q1. State the Binomial Theorem for positive integer index.

Answer: $(a+b)^n = \sum_{r=0}^n {^nC_r} a^{n-r} b^r$.

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Q2. How many terms are there in the expansion of $(a+b)^n$?

Answer: There are $n+1$ terms.

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Q3. Write the formula for the general term.

Answer: $T_{r+1} = {^nC_r} a^{n-r} b^r$.

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Q4. What is the sum of binomial coefficients ${^nC_0} + {^nC_1} + \dots + {^nC_n}$?

Answer: The sum is $2^n$.

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Q5. What is the coefficient of $x^n$ in $(1+x)^{2n}$?

Answer: It is $^{2n}C_n = \frac{(2n)!}{(n!)^2}$.