(B) Let the coefficients of two consecutive terms be $^nC_r$ and $^nC_{r+1}$.Given that $^nC_r = ^nC_{r+1}$.Using the formula $^nC_r = \frac{n!}{r!(n-

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

(B) Let the coefficients of two consecutive terms be $^nC_r$ and $^nC_{r+1}$.Given that $^nC_r = ^nC_{r+1}$.Using the formula $^nC_r = \frac{n!}{r!(n-r)!}$,we have:$\frac{n!}{r!(n-r)!} = \frac{n!}{(r+1)!(n-r-1)!}$$\frac{1}{r!(n-r)(n-r-1)!} = \frac{1}{(r+1)r!(n-r-1)!}$$\frac{1}{n-r} = \frac{1}{r+1}$$r + 1 = n - r$$n = 2r + 1$Since $r$ is an integer,$2r + 1$ is always an odd integer. Therefore,$n$ must be an odd integer.

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 coefficients of two consecutive terms in the expansion of $(1 + x)^n$ will be equal,if

A
$n$ is any integer
B
$n$ is an odd integer
C
$n$ is an even integer
D
None of these

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

The number of integral terms in the expansion of $\{7^{(1/2)} + 11^{(1/6)}\}^{824}$ is equal to ...................

DifficultJEE MAIN 2024

View Solution

Let $n$ be a positive even integer. If the ratio of the largest coefficient and the $2^{nd}$ largest coefficient in the expansion of $(1+x)^{n}$ is $11:10$, then the number of terms in the expansion of $(1+x)^{n}$ is:

EasyWBJEE 2013

View Solution

If the numerically greatest term in the expansion of $(2-3x)^9$ when $x=1$ is $P_1^\alpha P_2^\beta P_3^\gamma P_4^\delta$ (where $P_1 < P_2 < P_3 < P_4$ are the first four prime numbers),then $\alpha+\beta+\gamma+\delta=$

MediumAP EAMCET 2023

View Solution

The coefficient of $x^{4}$ in the expansion of $(1+x+x^{2}+x^{3})^{6}$ in powers of $x$ is:

MediumJEE MAIN 2020

View Solution

The ratio of the $5^{th}$ term from the beginning to the $5^{th}$ term from the end in the binomial expansion of $\left( 2^{1/3} + \frac{1}{2(3)^{1/3}} \right)^{10}$ is

DifficultJEE MAIN 2019

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 coefficients of two consecutive terms in the expansion of $(1 + x)^n$ will be equal,if

Similar Questions

The number of integral terms in the expansion of $\{7^{(1/2)} + 11^{(1/6)}\}^{824}$ is equal to ...................

Let $n$ be a positive even integer. If the ratio of the largest coefficient and the $2^{nd}$ largest coefficient in the expansion of $(1+x)^{n}$ is $11:10$, then the number of terms in the expansion of $(1+x)^{n}$ is:

If the numerically greatest term in the expansion of $(2-3x)^9$ when $x=1$ is $P_1^\alpha P_2^\beta P_3^\gamma P_4^\delta$ (where $P_1 < P_2 < P_3 < P_4$ are the first four prime numbers),then $\alpha+\beta+\gamma+\delta=$

The coefficient of $x^{4}$ in the expansion of $(1+x+x^{2}+x^{3})^{6}$ in powers of $x$ is:

The ratio of the $5^{th}$ term from the beginning to the $5^{th}$ term from the end in the binomial expansion of $\left( 2^{1/3} + \frac{1}{2(3)^{1/3}} \right)^{10}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module