# 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 ?

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,

$$(2\cos A)^4=(e^{iA}+e^{-iA})^4=2\cos4A+\binom412\cos2A+\binom42$$

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,

$$S=\frac{6n-5}{16}.$$

• Thank you both so much for your responses. A clever approach to this problem I would never have seen. May 13, 2019 at 12:19

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 $$\left( \begin{array}{cc} p & \sum\limits_{k=1}^{n} \cos^{2p}\Big(\frac{k\pi}{2n+1}\big) \\ 1 & \frac{2 n-1}{4} \\ 2 & \frac{6 n-5}{16} \\ 3 & \frac{10 n-11}{32} \\ 4 & \frac{70 n-93}{256} \\ 5 & \frac{126 n-193}{512} \\ 6 & \frac{462 n-793}{2048} \\ 7 & \frac{858 n-1619}{4096} \\ 8 & \frac{12870 n-26333}{65536} \end{array} \right)$$

• What a wonderful generalisation ! Thank you for expansion May 14, 2019 at 6:55