Check the identity $\sum\limits_{k=1}^n\sin^{2p}\left({k\pi\over 2n+1}\right)= 2^{-2p-1}{2p \choose p}(2n+1)$

Question

We want to evaluate the sums $$S_{n,p}=\sum_{k=1}^{n}\sin^{2p}\left({k\pi\over 2n+1}\right)$$ So we try many attempts by experimenting with the sum calculator which leads us to the result $$S_{n,p}={{2p \choose p}(2n+1)\over 2^{2p+1}}$$ How do we proof that the above result is correct or not?

Answer 1

Assuming that $p$ is a positive integer and letting $\omega = e^{\pi i/(2n+1)}$, we have

\begin{align*} \sin^{2p}\left( \frac{k\pi}{2n+1} \right) &= \frac{1}{2^{2p}} \sum_{j=0}^{2p} \binom{2p}{j} (-1)^{p-j} \omega^{2k(p-j)} \\ &= \frac{1}{2^{2p+1}} \sum_{j=0}^{2p} \binom{2p}{j} (-1)^{p-j} \left( \omega^{2k(p-j)} + \omega^{-2k(p-j)} \right). \end{align*}

Now utilizing the Dirichlet kernel formula, we obtain

$$ \sum_{k=1}^{n} \left( \omega^{2k(p-j)} + \omega^{-2k(p-j)} \right) = -1 + \sum_{k=-n}^{n} \omega^{2k(p-j)} = \begin{cases} 2n, & 2n+1 \mid p-j \\ -1, & 2n+1 \nmid p-j \end{cases}.$$

So it follows that

\begin{align*} \sum_{k=1}^{n} \sin^{2p}\left( \frac{k\pi}{2n+1} \right) &= \frac{1}{2^{2p+1}} \sum_{j=0}^{2p} \binom{2p}{j} (-1)^{p-j} \left( (2n+1)\mathbf{1}_{\{ 2n+1 \mid p-j \}} - 1 \right) \\ &= \frac{2n+1}{2^{2p+1}} \sum_{l} (-1)^{l} \binom{2p}{p+(2n+1)l}, \end{align*}

where the sum in the last line runs over all integers $l$ such that $|l| \leq \frac{p}{2n+1}$. In particular, if $p \leq 2n$ then only $l = 0$ contributes to the sum and we retrieve OP's formula.