(C) For the expansion of $(1 + x)^{2n}$,the total number of terms is $2n + 1$,which is odd.Therefore,there is only one middle term,which is the $\left

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(C) For the expansion of $(1 + x)^{2n}$,the total number of terms is $2n + 1$,which is odd.Therefore,there is only one middle term,which is the $\left(\frac{2n}{2} + 1\right)$-th term,i.e.,the $(n + 1)$-th term.The general term $T_{r+1}$ in the expansion of $(1 + x)^{2n}$ is given by $T_{r+1} = {}^{2n}C_r x^r$.For $r = n$,the middle term is $T_{n+1} = {}^{2n}C_n x^n$.Using the formula ${}^{2n}C_n = \frac{(2n)!}{n! n!}$,we get the middle term as $\frac{(2n)!}{(n!)^2} x^n$.

Class 11 Mathematics Binomial Theorem General term, Coefficient of any power of x, Independent term, Middle term and Greatest term and Greatest coefficient

Easy

The middle term in the expansion of $(1 + x)^{2n}$ is

A
$\frac{(2n)!}{n!} x^2$
B
$\frac{(2n)!}{n!(n - 1)!} x^{n + 1}$
C
$\frac{(2n)!}{(n!)^2} x^n$
D
$\frac{(2n)!}{(n + 1)!(n - 1)!} x^n$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Mathematics Questions

Chapter

Binomial Theorem

Subtopic

General term, Coefficient of any power of x, Independent term, Middle term and Greatest term and Greatest coefficient

In the expansion of $(1+x)^n$,the coefficients of the $p^{th}$ and $(p+1)^{th}$ terms are respectively $p$ and $q$. Then $p+q$ is equal to:

MediumTS EAMCET 2002

View Solution

If the $9^{th}$ and $10^{th}$ terms are the numerically greatest terms in the expansion of $(5x - 6y)^n$ when $x = 2/5$ and $y = 1/2$,then the absolute value of the middle term of that expansion is:

MediumTS EAMCET 2020

View Solution

Let the coefficients of three consecutive terms in the binomial expansion of $(1+2x)^n$ be in the ratio $2:5:8$. Then the coefficient of the term,which is in the middle of these three terms,is $...........$.

DifficultJEE MAIN 2023

View Solution

Find the middle term of the expansion of $ \left(\frac{10}{x}+\frac{x}{10}\right)^{10} $.

EasyKCET 2015

View Solution

If the coefficients of the middle terms in the binomial expansions of $(1+\alpha x)^{26}$ and $(1-\alpha x)^{28}$,where $\alpha \neq 0$,are equal,then the value of $\alpha$ is:

DifficultJEE MAIN 2026

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

The middle term in the expansion of $(1 + x)^{2n}$ is

Similar Questions

In the expansion of $(1+x)^n$,the coefficients of the $p^{th}$ and $(p+1)^{th}$ terms are respectively $p$ and $q$. Then $p+q$ is equal to:

If the $9^{th}$ and $10^{th}$ terms are the numerically greatest terms in the expansion of $(5x - 6y)^n$ when $x = 2/5$ and $y = 1/2$,then the absolute value of the middle term of that expansion is:

Let the coefficients of three consecutive terms in the binomial expansion of $(1+2x)^n$ be in the ratio $2:5:8$. Then the coefficient of the term,which is in the middle of these three terms,is $...........$.

Find the middle term of the expansion of $ \left(\frac{10}{x}+\frac{x}{10}\right)^{10} $.

If the coefficients of the middle terms in the binomial expansions of $(1+\alpha x)^{26}$ and $(1-\alpha x)^{28}$,where $\alpha \neq 0$,are equal,then the value of $\alpha$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module