Find the closed form of $\sum_{k=1}^{n}\cos^{2m+1}\left(\frac{(2k-1)\pi}{2n+1}\right)$ Add it:  Let $m,n$ be postive integers. Find the closed form of 
$$f=\sum_{k=1}^{n} \cos^{2m+1}{\left(\dfrac{(2k-1)\pi}{2n+1}\right)}$$
for $m, n \in \mathbb{N}^{+}$.
Maybe use Euler 
\begin{align}
2\cos{x} &=e^{ix}+e^{-ix} \\
\dfrac{\pi}{2n+1} &=x
\end{align}
then
\begin{align}
f &= \dfrac{1}{2^{2m+1}}\sum_{k=1}^{n}(w^{(2k-1)}+w^{-(2k-1)})^{2m+1} \\
&=\dfrac{1}{2^{2m+1}}\sum_{k=1}^{n}\sum_{i=0}^{2m+1}\binom{2m+1}{i}w^{-(2k-1)i}w^{((2m+1)-i)(2k-1)}\\
&=\dfrac{1}{(2w)^{2m+1}}\sum_{i=0}^{2m+1}\binom{2m+1}{i}w^{2i}\sum_{k=1}^{n}w^{(4m-4i+2)k}
\end{align}
where $w=e^{ix}$.
 A: The function $\frac{(2n+1)/z}{z^{2n+1}-1}$ has residue $1$ at each $2n+1^\text{st}$ root of unity. So we need to account for the residues at $0$ and $\infty$.
$$\newcommand{\Res}{\operatorname*{Res}}
\begin{align}
\sum_{k=1}^n\cos^{2m+1}\left(\pi\,\frac{2k-1}{2n+1}\right)
&=\frac12-\frac12\sum_{k=0}^{2n}\cos^{2m+1}\left(2\pi\,\frac{k}{2n+1}\right)\\
&=\frac12+\frac12\Res_{z=0}\left(\frac1{2^{2m+1}}\frac{(2n+1)/z}{z^{2n+1}-1}\left(z+\frac1z\right)^{2m+1}\right)\\
&-\frac12\Res_{z=\infty}\left(\frac1{2^{2m+1}}\frac{(2n+1)/z}{z^{2n+1}-1}\left(z+\frac1z\right)^{2m+1}\right)\\
&=\frac12-\frac{2n+1}{2^{2m+2}}\left[z^0\right]\left(\sum_{k=0}^\infty z^{(2n+1)k}\left(z+\frac1z\right)^{2m+1}\right)\\
&-\frac{2n+1}{2^{2m+2}}\left[z^0\right]\left(\sum_{k=1}^\infty z^{-(2n+1)k}\left(z+\frac1z\right)^{2m+1}\right)\\
&=\frac12-\frac{2n+1}{2^{2m+1}}\sum_{k=0}^m\binom{2m+1}{k}[\,2n+1\,|\,2m+1-2k\,]
\end{align}
$$
where $[\dots]$ are Iverson Brackets.
Note that for $m\lt n$, the sum is $\frac12$ because $2n+1$ cannot divide $2m+1-2k$.
