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

$T_{r+1}=C(2n,n)x^n$

Putting $r=n$ we have

$T_{n+1}=C(2n,n)x^n$

Coefficient of $x^n=C(2n,n)$

Again general term in the expansion of $(1+x)^{2n-1}$ is

$T_{r+1}=C(2n-1,r)x^r$

Putting $r=n$ we have

$T_{n+1}=C(2n-1,n)x^n$

Coefficient of $x^n$ in the expansion of $x^n$ is $C(2n-1,n)$

According to the problem we have to prove that

$C(2n,n)=2\times C(2n-1,n)$

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

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

Multiplying $N^r$ and $D^r$ by n on RHS we have

$\large\frac{2n(2n-1)!}{n!n(n-1)!}$

i.e $\large\frac{2n!}{n!n!}=\frac{2n!}{n!n!}$ which is true.

Hence proved.