Show that $\sum_{n=1}^{m}\sin^{2p}\left(\frac{n\pi}{2m+1}\right)=\frac{{2p\choose p}(2m+1)}{2^{2p+1}}$ Show that,
$(p,m)\ge 1$
$$\sum_{n=1}^{m}\sin^{2p}\left(\frac{n\pi}{2m+1}\right)=\frac{{2p\choose p}(2m+1)}{2^{2p+1}}$$
The central binomial coefficient is defined by ${2n\choose n}=\frac{(2n)!}{(n!)^2}$ for all $n\ge 0$
May be this would helpful, $\frac{2n}{(2n+1){2n\choose n}}=\int_0^1(2x(1-x))^ndx$
I don't know how to go about proving this identity, can anybody help me to prove it?
 A: The identity is true when $2m+1>p$. Let's write the sum in this way
$$S=\sum_{n=1}^m\sin^{2p} \left( \frac{\pi n}{2m+1}\right)=\frac{(-1)^p}{2^{2p}}\sum_{n=1}^m(e^{\frac{i\pi n}{2m+1}}-e^{-\frac{i\pi n}{2m+1}})^{2p}.$$
Now let's expand the powers
$$(e^{\frac{i\pi n}{2m+1}}-e^{-\frac{i\pi n}{2m+1}})^{2p}=
\sum_{k=0}^{2p} {2p \choose k} e^{\frac{2i\pi n}{2m+1}\cdot (p-k)}(-1)^k.$$
Notice that if $2m+1>p$ using the geometric sum we can sum over $n$ the $k$-th terms ($k \neq p$)
$$S(k)=\sum_{n=1}^{m}e^{\frac{2i\pi n}{2m+1}\cdot (p-k)}=\frac{1-e^{\frac{2i\pi (m+1)}{2m+1}\cdot (p-k)}}{1-e^{\frac{2i\pi}{2m+1}\cdot (p-k)}}-1$$
the sum of the $2p-k$-terms is similar
$$S(2p-k)=\sum_{n=1}^{m}e^{\frac{2i\pi n}{2m+1}\cdot (k-p)}=\frac{1-e^{\frac{2i\pi (m+1)}{2m+1}\cdot (k-p)}}{1-e^{\frac{2i\pi}{2m+1}\cdot (k-p)}}-1$$
Summing those two we see that everything cancels and
$$S(k)+S(2p-k)=-1.$$
The sum over $n$ of the $p-th$ term is just $m$. Hence we can rewrite the original sum in this form
$$S=\frac{(-1)^p}{2^{2p}} \left ( \sum_{k=0}^{p-1}(-1)^{k+1}{2p \choose k}+m {2p \choose p}(-1)^p\right)=\color{red} {\frac{{2p \choose p}(2m+1)}{2^{2p+1}}}$$
We used the fact that 
 $$-\sum_{k=0}^{p-1}(-1)^{k}{2p \choose k}={2p \choose p}\frac{(-1)^p}{2}.$$
