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

The general term $(T_{r+1})$ in the binomial expansion of $(a+b)^{m}$ is given by $T_{r+1} = {}^{m}C_{r} a^{m-r} b^{r}$.
For the expansion of $(1+x)^{2n}$,the coefficient of $x^{n}$ is obtained by setting $r=n$:
Coefficient $= {}^{2n}C_{n} = \frac{(2n)!}{n!(2n-n)!} = \frac{(2n)!}{(n!)^2}$ ........... $(1)$
For the expansion of $(1+x)^{2n-1}$,the coefficient of $x^{n}$ is obtained by setting $r=n$:
Coefficient $= {}^{2n-1}C_{n} = \frac{(2n-1)!}{n!(2n-1-n)!} = \frac{(2n-1)!}{n!(n-1)!}$
Multiply the numerator and denominator by $2n$:
$= \frac{2n \cdot (2n-1)!}{2n \cdot n!(n-1)!} = \frac{(2n)!}{2 \cdot n! \cdot n!} = \frac{1}{2} \left[ \frac{(2n)!}{(n!)^2} \right]$ ........... $(2)$
Comparing $(1)$ and $(2)$:
${}^{2n}C_{n} = 2 \cdot {}^{2n-1}C_{n}$
Thus,the coefficient of $x^{n}$ in $(1+x)^{2n}$ is twice the coefficient of $x^{n}$ in $(1+x)^{2n-1}$.