Question

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 $(1+x{)}^{2n-1}$

• A. True
• B. False

Answer 1

The correct option is A True

$(1+x{)}^{2n}={}^{2n}{C}_0{+}^{2n}{C}_1{x}^1{+}^{2n}{C}_2{x}^2+......{+}^{2n}{C}_n{x}^n$

$(1+x{)}^{2n-1}={}^{2n-1}{C}_0{+}^{2n-1}{C}_1{x}^1{+}^{2n-1}{C}_2{x}^2+.....{+}^{2n-1}{C}_n{x}^n$

Coefficient of $x^n$ $in(1+x{)}^{2n-1}is{(}^{2n-1}{C}_n)$

${}^{2n}{C}_n=\frac{\left(2n\right)!}{n!n!}$

$=\frac{\left(2n\right)(2n-1)(2n-2)!}{n(n-1)!n!}$

$=\frac{\left(2n\right)(2n-1)2(n-1)!}{n(n-1)!n!}$

$=2\left[\frac{2(2n-1)!}{n!}\right]............\left(i\right)$

Now, ${}^{2n-1}{C}_n=\frac{(2n-1)!}{n!\left(2n-1-n\right)!}$

$=\frac{(2n-1)!}{n!\left(n-1\right)!}$

$=\frac{(2n-1)(2n-2)!}{n!\left(n-1\right)!}$

$=\frac{(2n-1)2(n-1)!}{n!\left(n-1\right)!}$

$=\frac{2(2n-1)}{n!}.............\left(ii\right)$

From (i) and (ii), we have

${}^{2n}{C}_n={2.}^{2n-1}{C}_n$

Hence, it is true.