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# General and Middle Terms - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• General Term of Binomial Expansion is considered one the most difficult concept.

• Middle Term is considered one of the most asked concept.

• 301 Questions around this concept.

## Solve by difficulty

In the binomial expansion of $\dpi{100} (a-b)^{n},n\geq 5,$ the sum of 5th and 6th terms is zero, then $\dpi{100} a/b$ equals:

What is the 5th term in the expansion of $\left ( 3a+b \right )^9?$

coefficient of x5 in the expansion of $(1+x^2)^5 (1+x)^2$  is

The term that is independent of x, in the expression $\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9$ is:

The term independent of x in the expansion of $\left(\frac{3 x^2}{2}-\frac{1}{3 x}\right)^9 is$

The coefficient of a4b5 in the expansion of (a + b)9 is

In the expansion of $\left(x^3-\frac{1}{x^2}\right)^{15}$, the constant term is:

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If the coefficients of the $r^t^h$ and $(r+1)^t^h$ terms in the expansion of $(3+7 x)^{29}$ are equal, then $r$ is equal to

Find the coefficient of $s^{12} t^{13}$  in the expansion of $(2 s-3 t)^{25}$.

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The term independent of x in $\left[\sqrt{\frac{x}{3}}+\sqrt{\left(\frac{3}{2 x^2}\right)}\right]^{10}$ is

## Concepts Covered - 6

General Term of Binomial Expansion

General Term

Sometimes we are interested only in a certain term of a binomial expansion. We do not need to fully expand a binomial to find a single specific term.

$\left(r+1\right)^{\mathrm{th}} \text { term is called as general term in }(x+y)^{n}\;\text{and general term is given by}$

$\mathrm{T}_{\mathrm{r}+1}=^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{\;x}^{\mathrm{n}-\mathrm{r}} \cdot \mathrm{y}^{\mathrm{r}}$

Term independent of x

It means term containing x0,

For example, to find term independent of x in $\left(x-\frac{1}{x}\right)^{20}$

$\mathrm{\begin{array}{l}{\left(x-\frac{1}{x}\right)^{20}} \\ {\Rightarrow \quad T_{r+1}=^{20} C_{r} x^{20-r}(-1)^{r} \frac{1}{x^{r}}=20 C_{r} x^{20-2 r}(-1)^r} \\ {\Rightarrow \quad 20-2 r=0 ; r=10} \\ {\Rightarrow \quad 11^{t h} \text { term is independent of } x}\end{array}}$

(p+1)th term from the End

(p+1)th term from the end

The binomial expansion $\mathrm{(x+y)^{n}=^{n} \mathrm{C}_{0} x^{n}+^{n} \mathrm{C}_{1} x^{n-1} y+^{n} \mathrm{C}_{2} x^{n-2} y^{2}+\cdots+^{n} \mathrm{C}_{n} y^{n}}$

From Starting

$\\\mathrm{ \underbrace{^{n} \mathrm{C}_{0} \;x^{n}}+\underbrace{^{n} \mathrm{C}_{1}\; x^{n-1} \;y}+^{n} \mathrm{C}_{2}\; x^{n-2} \;y^{2}+\cdots+\underbrace{^nC_{n-1}\;x\;y^{n-1}}+\;\underbrace{^{n} \mathrm{C}_{n} \;y^{n}}}\\ {\color{DarkGreen} \boldsymbol{\mathbf{\;\;\;1^{st}}}}\;\;\;\;\;\;\;\;\;{\color{DarkBlue} \boldsymbol{\mathbf{\;\;\;2^{nd}}}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\color{DarkRed} \boldsymbol{\mathbf{\;\;\;n^{th}}}}\;\;\;\;\;{\color{Teal} \boldsymbol{\mathbf{\;\;\;\;(n+1)^{th}}}}\\ {\color{DarkGreen} \boldsymbol{\mathbf{\;term}}}\;\;\;\;\;\;\;{\color{DarkBlue} \boldsymbol{\mathbf{\;term}}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\color{DarkRed} \boldsymbol{\mathbf{\;term}}}\;\;\;\;\;{\color{Teal} \boldsymbol{\mathbf{\;\;\;\;\;term}}}$

From the End

$\\\\\mathrm{ \underbrace{^{n} \mathrm{C}_{0} \;y^{n}}+\underbrace{^{n} \mathrm{C}_{1}\; y^{n-1} \;x}+^{n} \mathrm{C}_{2}\; y^{n-2} \;x^{2}+\cdots+\underbrace{^nC_{n-1}\;y\;x^{n-1}}+\;\underbrace{^{n} \mathrm{C}_{n} \;x^{n}}}\\ {\color{DarkGreen} \boldsymbol{\mathbf{\;\;\;1^{st}}}}\;\;\;\;\;\;\;\;\;{\color{DarkBlue} \boldsymbol{\mathbf{\;\;\;2^{nd}}}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\color{DarkRed} \boldsymbol{\mathbf{\;\;\;n^{th}}}}\;\;\;\;\;{\color{Teal} \boldsymbol{\mathbf{\;\;\;\;(n+1)^{th}}}}\\ {\color{DarkGreen} \boldsymbol{\mathbf{\;term}}}\;\;\;\;\;\;\;{\color{DarkBlue} \boldsymbol{\mathbf{\;term}}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\color{DarkRed} \boldsymbol{\mathbf{\;term}}}\;\;\;\;\;{\color{Teal} \boldsymbol{\mathbf{\;\;\;\;\;term}}}$

(Using relation $^n C_r = ^n C_{(n-r)}$ )

Now,

Consider the binomial expansion

$\mathrm{(y+x)^{n}=^{n} \mathrm{C}_{0} y^{n}+^{n} \mathrm{C}_{1} y^{n-1} x+^{n} \mathrm{C}_{2} y^{n-2} x^{2}+\cdots+^{n} \mathrm{C}_{n} x^{n}}$

Just observe that, (p + 1)th term from the end of the expansion of (x + y)n = (p + 1)th term from the beginning of the expansion of

(y + x)n = nCp yn-p xp

Radical Free terms or Rational Terms

Rational term in the expansion of $\mathrm{(x^{1/a}+y^{1/b})^N,\;x,y\in\;prime\;numbers} .$

$\begin{array}{l}{\text { First, find } T_{r+1}=^{N} C_{r}\left(x^{1 / a}\right)^{N-r}\left(y^{1 / b}\right)^{r}} \\ {\therefore \quad T_{r+1}=^{N} C_{r} \cdot x^{(N-r) / a} \cdot y^{r / b}} \\ {\text { By observation,} \text { when }} {\text { indices of } x \text { and } y \text { are integers, then the entire term will be rational }}\end{array}$

For example,

Find the number of terms in the expansion of  $(\sqrt[4]{9}+\sqrt[6]{8})^{100}$  which are rational

To make x and y as prime numbers, we can rewrite the expression as

$(\sqrt[4]{9}+\sqrt[6]{8})^{100}=\left(9^{1 / 4}+8^{1 / 6}\right)^{100}=\left(3^{1 / 2}+2^{1 / 2}\right)^{100}$\begin{aligned} \therefore \text { General term, } T_{r+1} &=^{100} C_{r}\left(3^{1 / 2}\right)^{100-r} \cdot\left(2^{1 / 2}\right)^{r} \\ &=^{100} C_{r} \cdot 3^{\frac{100-r}{2}} \cdot 2^{r / 2} \\ &=^{100} C_{r} \cdot 3^{50-r / 2} \cdot 2^{r / 2} \end{aligned}

$\begin{array}{l}{\text { Now, } \quad 0 \leq r \leq 100} \\ {\text { For } r=0,2,4,6,8, \ldots, 100, \text { indices of } 3 \text { and } 2 \text { are positive integers. }} \\ {\text { Hence, number of terms which are rational }=50+1=51}\end{array}$

Middle Term

The middle term in the expansion (x + y)n, depends on the value of 'n'.

Case 1 When 'n' is even

If n is even, and number of terms in the expansion is n + 1, so n +1 is odd number therefore only one middle term is obtained which is

$\left(\frac{\mathrm{n}}{2}+1\right)^{\mathrm{th}}$term.

It is given by

$\mathrm{T} _{\frac{\mathrm{n}}{2}+1}=\left(\begin{array}{c}{\mathrm{n}} \\ {\frac{\mathrm{n}}{2}}\end{array}\right) \mathrm{x}^{\frac{\mathrm{n}}{2}} \mathrm{y}^{\frac{\mathrm{n}}{2}}$

Case 2 When 'n' is odd

In this case, the number of terms in the expansion will be n + 1. Since n is odd so, n + 1 is even. Therefore, there will be two middle terms in the expansion, namely$\left(\frac{n+1}{2}\right)^{t h} \text { and }\left(\frac{n+3}{2}\right)^{t h}$ terms.

And it is given by

$\mathrm{T}_{\frac{\mathrm{n}+1}{2}}=\left(\begin{array}{c}{\mathrm{n}} \\ {\frac{\mathrm{n}-1}{2}}\end{array}\right) \mathrm{x}^{\frac{\mathrm{n}+1}{2}} \cdot \mathrm{y}^{\frac{\mathrm{n}-1}{2}} \,\,\,and \quad \mathrm{T}_{\frac{\mathrm{n}+3}{2}}=\left(\begin{array}{c}{\mathrm{n}} \\ {\frac{\mathrm{n}+1}{2}}\end{array}\right) \mathrm{x}^{\frac{\mathrm{n}-1}{2}} \cdot \mathrm{y}^{\frac{\mathrm{n}+1}{2}}$

Note:

Binomial Coefficient of the middle term is greatest among all binomial coefficients in an expansion.

• So if n is even, then nCis largest if r = n/2
• And if n is odd, then nCis largest if r = $\frac{n-1}{2}$ or $\frac{n+1}{2}$, and both these values of nC are equal
Consecutive coefficients of binomial Expansion

If consecutive coefficients are given

If in the question, coefficients of consecutive terms, like rth, (r + 1)th and (r + 2)th and so on… are involed, then to solve divide consecutive coefficients pairwise.

i.e.$\mathrm{\frac{\text{coefficient of (r+1)th of term }}{\text{coefficient of (r)th of term }},\;\frac{\text{coefficient of (r+2)th of term }}{\text{coefficient of (r+1)th of term }}....}$

And with the help of the given condition in the question, you get equations. Solve these equations to get the answer.

Let’s understand this by an example,

If the coefficients of rth, (r + 1)th and (r + 2)th terms of the expansion (1 + x)n  are in AP, and the relation between n and r is asked (where n  is a positive integer), then

From the concept of the general term

\begin{aligned} T_{r} &=T_{(r-1)+1}=^{n} C_{r-1} x^{r-1} \\ T_{r+1} &=^{n} C_{r} x^{r} \text { and } T_{r+2}=T_{(r+1)+1}=^{n} C_{r+1} x^{r+1} \end{aligned}

$\begin{array}{l}{\therefore \text { Coefficients of } r \text { th, }(r+1) \text { th and }(r+2) \text { th terms in the }} \\ {\text { expansion of }} \\ {\qquad(1+x)^{n} \text { are }^{n} C_{r-1},^{n} C_{r},^{n} C_{r+1}} \\ {\because \text { Given, }^{n} C_{r-1},^{n} C_{r},^{n} C_{r+1} \text { are in AP. }} \\ {\text { and } \quad n \geq r+1}\end{array}$

$\\{\therefore \frac{^{n} C_{r-1}}{^{n} C_{r}}, 1, \frac{^{n} C_{r+1}}{^{n} C_{r}} \text { are also in AP. }} \\ {\Rightarrow \frac{r}{n-r+1}, 1, \frac{n-r}{r+1} \text { are in AP. }}$

$\Rightarrow 1-\frac{r}{n-r+1}=\frac{n-r}{r+1}-1 \Rightarrow \frac{n-2 r+1}{n-r+1}=\frac{n-2 r-1}{r+1}$

\begin{aligned} \Rightarrow n r-2 r^{2}+r &+n-2 r+1 =n^{2}-2 n r-n-n r+2 r^{2}+r+n-2 r-1 \end{aligned}

$\Rightarrow \quad n^{2}-4 n r+4 r^{2}=n+2 \Rightarrow(n-2 r)^{2}=n+2$

Consecutive Terms of Binomial Expansion

If consecutive terms are given

If in the question, consecutive terms Tr , Tr + 1 , Tr + 2 ,  and so on…. are given. Then to divide consecutive coefficients pairwise i.e.$\mathrm{\frac{T_{r+1}}{T_r}=\alpha_1,\frac{T_{r+2}}{T_{r+1}}=\alpha_2,\;\frac{T_{r+3}}{T_{r+2}}=\alpha_3..........}$

Now divide α2 by α1 and α3 by α2 ...to solve the question

## Study it with Videos

General Term of Binomial Expansion
(p+1)th term from the End
Radical Free terms or Rational Terms

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