General term in the expansion of $(1+x)^{2 n}$ is
$T_{r+1}=C(2 n, r) x^r$
Putting $r=n$, we have
$T_{n+1}=C(2 n, n) x^n$
Coefficient of $x^n=C(2 n, n)$
Again general term in the expansion of
$(1+x)^{2 n-1} \text { is } T_{r+1}=C(2 n-1, r) x^r$
Putting $r=n$, we have
$T_{n+1}=C(2 n-1, n) x^n$
Coefficient of $x^n$ in the expansion of $(1+x)^{2 n-1}$ is $C(2 n-1, n)$
According to the problem, we have to prove that
$C(2 n, n)=2 \times C(2 n-1, n)$
or $\frac{2 n !}{n !(2 n-n) !}=2 \cdot \frac{(2 n-1) !}{n !(2 n-1-n) !}$
or $\frac{2 n !}{n ! n !}=2 \cdot \frac{(2 n-1) !}{n !(n-1) !}$
Multiplying $N^r$ and $D^r$ by $n$ on RHS, we have
$\frac{2 n !}{n ! n !}=\frac{2 n(2 n-1) !}{n ! n(n-1) !}$
i.e $\frac{2 n !}{n ! n !}=\frac{2 n !}{n ! n !}$, Which is true.
Hence proved.