Show that the coefficient of the middle term in the expansion of $(1+x)^{2n}$ is equal to the sum of the coefficients of the two middle terms in the expansion of $(1+x)^{2n-1}$.

(N/A) Since $2n$ is even,the expansion of $(1+x)^{2n}$ has only one middle term,which is the $(\frac{2n}{2}+1)^{\text{th}}$ term,i.e.,the $(n+1)^{\text{th}}$ term.
The $(n+1)^{\text{th}}$ term is given by $^{2n}C_{n}x^{n}$. Thus,the coefficient of $x^{n}$ is $^{2n}C_{n}$.
Similarly,since $(2n-1)$ is odd,the expansion of $(1+x)^{2n-1}$ has two middle terms,which are the $(\frac{2n-1+1}{2})^{\text{th}}$ and $(\frac{2n-1+1}{2}+1)^{\text{th}}$ terms,i.e.,the $n^{\text{th}}$ and $(n+1)^{\text{th}}$ terms.
The coefficients of these terms are $^{2n-1}C_{n-1}$ and $^{2n-1}C_{n}$,respectively.
Using the identity $^{n}C_{r-1} + ^{n}C_{r} = ^{n+1}C_{r}$,we have:
$^{2n-1}C_{n-1} + ^{2n-1}C_{n} = ^{2n}C_{n}$.
This proves that the coefficient of the middle term in $(1+x)^{2n}$ is equal to the sum of the coefficients of the two middle terms in $(1+x)^{2n-1}$.