Sum of Fourth Powers of cosine series has closed form solution. A problem posed in the 1988 Irish Mathematical Olympiad asks for a proof of the following
$$\sum\limits_{k=1}^{n} \cos^{4}\Big(\frac{k\pi}{2n+1}\big) = \frac{6n-5}{16}$$ 
Can anyone give me a heads-up on how to proceed to prove this very interesting result ?
 A: Hint:
$$\cos2A=2\cos^2A-1\iff2\cos^2A=1+\cos2A$$
$$\implies(2\cos^2A)^2=(1+\cos2A)^2=1+2\cos2A+\cos^22A$$
$$\implies8\cos^4A=2+4\cos2A+1+\cos4A$$
Alternatively, 
using How to prove Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$?,
$$(2\cos A)^4=(e^{iA}+e^{-iA})^4=2\cos4A+\binom412\cos2A+\binom42$$
Finally, use
How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?
A: Let $z=e^{i\pi/(2n+1)}$, and
 we have $$\left(\frac{z+z^{-1}}2\right)^4=\frac{z^4+4z^2+6+4z^{-2}+z^{-4}}{16}.$$
Now, summing for $z^k$ with $k$ in $[0,n]$ and using the geometric summation formula,
$$16S:=\frac{z^{4n+1}-1}{z-1}+4\frac{z^{2n+1}-1}{z-1}+6n+4\frac{z^{-2n-1}-1}{z^{-1}-1}+\frac{z^{-4n-1}-1}{z^{-1}-1}.$$
As $z^{2n+1}=-1$, this simplifies to
$$16S:=\frac{z^{-1}-1}{z-1}+4\frac{-1-1}{z-1}+6n-4z\frac{-1-1}{z-1}-\frac{z^{-1}-1}{z-1}=6n-4,$$ from which we deduct $1$ because of the term for $k=0$.
Finally,
$$S=\frac{6n-5}{16}.$$
A: Using the hints given in answers, we can make the result more general
$$\color{blue}{\sum\limits_{k=1}^{n} \cos^{2p}\Big(\frac{k\pi}{2n+1}\big) =  4^{-p} \binom{2 p}{p}\,n+4^{-p} \binom{2 p-1}{p}-\frac{1}{2}}$$ and get the beautiful
$$\left(
\begin{array}{cc}
p & \sum\limits_{k=1}^{n} \cos^{2p}\Big(\frac{k\pi}{2n+1}\big) \\
 1 & \frac{2 n-1}{4} \\
 2 & \frac{6 n-5}{16}  \\
 3 & \frac{10 n-11}{32}  \\
 4 & \frac{70 n-93}{256}  \\
 5 & \frac{126 n-193}{512}  \\
 6 & \frac{462 n-793}{2048} \\
 7 & \frac{858 n-1619}{4096} \\
 8 & \frac{12870 n-26333}{65536}
\end{array}
\right)$$
