General term and specific term

A typical JEE question asks: find the coefficient of $x^7$ in $(2x - 3/x)^{15}$ . The answer comes from one formula , the general term.

The formula

The general term in the expansion of $(a + b)^n$ is the $(r + 1)^{\text{th}}$ term: $\boxed{T_{r+1} = \binom{n}{r}\,a^{n - r}\,b^r, \quad r = 0, 1, 2, \dots, n.}$

So $T_1 = \binom{n}{0} a^n$ , $T_2 = \binom{n}{1} a^{n-1} b$ , $\dots$ , $T_{n+1} = \binom{n}{n} b^n$ .

Three standard questions

(a) Find the $k$ -th term. Set $r = k - 1$ in $T_{r+1}$ .

(b) Find the term containing $x^k$ . Simplify $T_{r+1}$ , set the exponent of $x$ to $k$ , solve for $r$ .

(c) Find the term independent of $x$ (i.e., constant term). Set the exponent of $x$ to $0$ .

Worked examples

Example 1. Find the $5^{\text{th}}$ term in the expansion of $(x + 2)^7$ .

$T_{r+1} = \binom{7}{r} x^{7-r} \cdot 2^r$ . For $T_5$ , $r = 4$ : $T_5 = \binom{7}{4} x^3 \cdot 2^4 = 35 \cdot 16 \cdot x^3 = 560 x^3$ .

Example 2. Find the term containing $x^9$ in $\left(2x - \dfrac{1}{x}\right)^{15}$ .

$T_{r+1} = \binom{15}{r} (2x)^{15-r} \left(-\dfrac{1}{x}\right)^r = \binom{15}{r} 2^{15-r} (-1)^r x^{15 - r} \cdot x^{-r} = \binom{15}{r} 2^{15-r} (-1)^r x^{15 - 2r}$ .

Set $15 - 2r = 9 \Rightarrow r = 3$ . Then the term is $\binom{15}{3} 2^{12} (-1)^3 x^9 = 455 \cdot 4096 \cdot (-1) \cdot x^9 = -1{,}863{,}680\,x^9$ .

Example 3. Find the term independent of $x$ in $\left(x^2 + \dfrac{1}{x}\right)^9$ .

$T_{r+1} = \binom{9}{r} (x^2)^{9-r} \left(\dfrac{1}{x}\right)^r = \binom{9}{r} x^{18 - 2r - r} = \binom{9}{r} x^{18 - 3r}$ .

Constant: $18 - 3r = 0 \Rightarrow r = 6$ . Term: $\binom{9}{6} = 84$ .

Example 4. Find the coefficient of $x^5$ in $\left(\dfrac{x}{2} - \dfrac{3}{x^2}\right)^{11}$ .

$T_{r+1} = \binom{11}{r} \left(\dfrac{x}{2}\right)^{11-r} \left(-\dfrac{3}{x^2}\right)^r = \binom{11}{r} \dfrac{x^{11-r}}{2^{11-r}} \cdot (-3)^r x^{-2r} = \binom{11}{r} \dfrac{(-3)^r}{2^{11-r}} x^{11 - 3r}$ .

Set $11 - 3r = 5 \Rightarrow r = 2$ . Coefficient: $\binom{11}{2} \dfrac{9}{2^9} = 55 \cdot \dfrac{9}{512} = \dfrac{495}{512}$ .

Example 5 (harder). Find the ratio of $T_4$ to $T_3$ in $(1 + x)^n$ if $T_3 = 28 x^2$ and $n$ is a positive integer.

$T_3 = \binom{n}{2} x^2 = 28 x^2 \Rightarrow \binom{n}{2} = 28 \Rightarrow n = 8$ .

$\dfrac{T_4}{T_3} = \dfrac{\binom{n}{3} x^3}{\binom{n}{2} x^2} = \dfrac{n - 2}{3} x = \dfrac{6}{3} x = 2x$ .

Try it yourself

Find the $4^{\text{th}}$ term in $(2x + 1)^6$ .
Find the term containing $x^7$ in $(x^2 + 1)^{10}$ .
Find the constant term in $\left(x - \dfrac{1}{x}\right)^{10}$ .
Find the coefficient of $x^6$ in $\left(2x^2 - \dfrac{1}{x}\right)^9$ .
Find the $6^{\text{th}}$ term in $\left(2 - \dfrac{x}{3}\right)^{10}$ .
If the third term in $(1 + x)^n$ is $45 x^2$ , find $n$ .
Find the term independent of $x$ in $\left(\sqrt{x} + \dfrac{1}{x^2}\right)^{10}$ .
Find the term containing $x^{-5}$ in $\left(x + \dfrac{2}{x^2}\right)^{10}$ .
Find the $7^{\text{th}}$ term in the expansion of $(1 - 2x)^{12}$ .
Compute the ratio $T_5 / T_4$ in the expansion of $(1 + x)^{10}$ at $x = 1$ .
Find the coefficient of $x^4$ in $\left(\dfrac{x^2}{2} - \dfrac{1}{x}\right)^{10}$ .
If the coefficients of $T_5$ and $T_8$ in $(1 + x)^n$ are equal, find $n$ .

Pitfalls / Tricks

$T_{r+1}$ is the $(r+1)$ -th term, not the $r$ -th. So $T_1$ corresponds to $r = 0$ .
Always factor out the negative signs first when one or both of $a, b$ contain a $-$ .
Set the exponent of $x$ in $T_{r+1}$ to the desired value, then solve for $r$ . If $r$ is not an integer between $0$ and $n$ , the term does not exist.
Insight. The general term is just one formula. Memorise $T_{r+1} = \binom{n}{r} a^{n-r} b^r$ , every problem in this subtopic reduces to choosing the right $r$ .