Question

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

Answer 1

In $(1+x{)}^{2n}$, we have $T_{r+1}{=}^{2n}{C}_r.x^r.$

$\therefore$ Coefficient of $x^n\text{in}(1+x{)}^{2n}{=}^{2n}{C}_n=\frac{\left(2n\right)!}{(n!)(n!)}=\frac{\left(2n\right)(2n-1)!}{n(n-1)!\times (n!)}$

$=2\times \frac{(2n-1)!}{(n-1)!\times (n!)}.$

In $(1+x{)}^{2n-1}$, we have $T_{r+1}{=}^{2n-1}{C}_r{x}^r.$

$\therefore$ coefficient of $x^n$ in $(1+x{)}^{2n-1}{=}^{2n-1}{C}_n=\frac{(2n-1)!}{(2n-1-n)(n-1)!}=\frac{(2n-1)!}{(n-1)!\times n!}$

Hence, coefficient of $x^n\text{in}(1+x{)}^{2n}=2\times$ coeffcient of $x^n$ in $(1+x{)}^{2n-1}$