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

It is known that $(r+1)^{\text {th }}$ term, $\left(T_{r+1}\right),$ in the binomial expansion of $(a+b)^{n}$ is given by

${T_{r + 1}} = {\,^n}{C_r}{a^{n - r}}{b^r}$

Assuming that $x^{n}$ occurs in the $(r+1)^{\text {th }}$ term of the expansion of $(1+x)^{2 n}$, we obtain

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

Comparing the indices of $x$ in $x^{n}$ and in $T_{r+1},$ we obtain $r=n$

Therefore, the coefficient of $x^{n}$ in the expansion of $(1+x)^{2 n}$ is

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

Assuming that $x^{n}$ occurs in the $(k+1)^{\text {th }}$ term of the expansion of $(1+x)^{2 n-1}$, we obtain

${T_{k + 1}} = {\,^{2n}}{C_k}{(1)^{2n - 1 - k}}{(x)^k} = {\,^{2n}}{C_k}{(x)^k}$

Comparing the indices of $x$ in $x^{n}$ and in $T_{k+1},$ we obtain $k=n$

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

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

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

From $(1)$ and $(2),$ it is observed that

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

$ \Rightarrow {\,^{2n}}{C_n} = 2{\rm{(}}{\,^{2n - 1}}{C_n}{\rm{)}}$

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

Hence proved.