Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Q11. Prove that the coefficient of $x^n$ in the expansion of $(1+x)^{2n}$ is twice the coefficient of $x^n$ in the expansion of $(1+x)^{2n-1}$ .

Answers (1)

best_answer

As we know that the general $(r+1)^{th}$ term $T_{r+1}$ in the binomial expansion of $(a+b)^n$ is given by

$T_{r+1}=^nC_ra^{n-r}b^r$

So, general $(r+1)^{th}$ term $T_{r+1}$ in the binomial expansion of $(1+x)^{2n}$ is,

$T_{r+1}=^{2n}C_r1^{2n-r}x^r$

$x^n$ will come when $r=n$ ,

So, Coefficient of $x^n$ in the binomial expansion of $(1+x)^{2n}$ is,

$K_{1x^n}=^{2n}C_n$

Now,

the general $(r+1)^{th}$ term $T_{r+1}$ in the binomial expansion of $(1+x)^{2n-1}$ is,

$T_{r+1}=^{2n-1}C_r1^{2n-1-r}x^r$

Here also $x^n$ will come when $r=n$ ,

So, Coefficient of $x^n$ in the binomial expansion of $(1+x)^{2n-1}$ is,

$K_{2x^n}=^{2n-1}C_n$

Now, As we can see

$^{2n-1}C_n=\frac{(2n-1)!}{n!(2n-1-n)!}=\frac{(2n-1)!}{n!(n-1)!}=\frac{(2n)!}{2n(n!)(n-1)!}=\frac{(2n)!}{2(n!)(n!)}$

$^{2n-1}C_n=\frac{1}{2}\times^{2n}C_n$

$2\times^{2n-1}C_n=^{2n}C_n$

$2\times K_{2x^n}=K_{1x^n}$

Hence, the coefficient of $x^n$ in the expansion of $(1+x)^{2n}$ is twice the coefficient of $x^n$ in the expansion of $(1+x)^{2n-1}$ .

Posted by

Pankaj Sanodiya

View full answer

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE board exams 2025 from February 15; Class 10, 12 subject-wise marks distribution

anam-khan

CBSE Date Sheet 2025: Class 10, 12 board exam dates soon on cbse.gov.in; previous trends, sample papers

anam-khan

CBSE Board Exams 2025: 75% attendance must for students to be eligible, reiterates board

Latest Question