Values of the sums $\sum\limits_{k=1}^{n}\cos^4(πk/(2n+1))$

Question

I have been given a question which asks you to prove that $$ \sum_{k=1}^{n}\cos^4\left(\frac{πk}{2n+1}\right)=\frac{6n-5}{16} $$

The main problem I have with solving this is that since the summands contain $n$, so $ \sum\limits_{k=1}^{n}\cos^4(\frac{πκ}{2n+1})$ is not the sum of constant terms. A proof by induction fails also.

I've tried generalising the problem by letting $2n+1=p$, where p is any positive integer. Then I replaced $\cos^4(x)$ with ${\frac{1+\cos2x}{2}}^2$, then finally simplifying to $\frac{3}{8}+\frac{\cos 2x}{2}+\frac{\cos 4x}{8}$.

We see that $ \sum \frac{3}{8} =\frac{3n}{8}$, which is the easy part.

Now for $\sum \frac{\cos(\frac{2kπ}{p})}{2}$, I tried switching this to the sum of $\frac{e^{\frac{2πki}{p}}+e^{\frac{-2πki}{p}}}{2}$. I noticed that since $e^{2πki}=1$ for positive integer $k$, the summand becomes $\frac{ω+\bar{ω}}{2}$ where $ω^p=1$. But I feel I have done something wrong here. The way shown above does not appear to lead anywhere, and I used a similar argument for $\frac{\cos4x}{8}$, without success.

How should I try solving a sum like this, where the summand contains a function of $n$?

Answer 1

HINT: Consider the sum $$ \sum_{k=0}^n \cos k\alpha + i\sin k\alpha=\sum_{k=0}^n e^{ik\alpha}. $$ The real part will be the sum of $\cos$'s.

Answer 2

Hint: $$\sum_{k=1}^{n}\cos^{4}\left(\frac{\pi k}{2n+1}\right)=\frac{1}{16}\sum_{k=1}^{n}\left(e^{-\pi ik/\left(2n+1\right)}+e^{\pi ik/\left(2n+1\right)}\right)^{4}= $$ $$=\frac{1}{16}\sum_{k=1}^{n}\left(4e^{-2i\pi k/\left(2n+1\right)}+4e^{2i\pi k/\left(2n+1\right)}+e^{-4i\pi k/\left(2n+1\right)}+e^{4i\pi k/\left(2n+1\right)}+6\right). $$

Answer 3

Put $z:=e^{i\pi/(2n+1)}$. The points $$z_k:=z^k\qquad(1\leq k\leq 4n+2)$$ are the vertices of a regular $(4n+2)$-gon $P$ inscribed in the unit circle. Denote by $\phi_k:={\rm arg}(z_k)$ the polar angles of these vertices. We then are told to compute $$S:=\sum_{k=1}^n\cos^4\phi_k={1\over4}\left(\sum_{k=1}^{4n+2}\cos^4\phi_k \ -2\right)\ ,$$ whereby we have made use of symmetry (note that the points $\pm1$ are vertices of $P$, but their contribution to the large sum is excluded from $S$). Now $\cos\phi_k={1\over2}(z_k+\bar z_k)$, and this gives $$\cos^4\phi_k={1\over16}\bigl(z_k^4+4z_k^2 +6+4\bar z_k^2+\bar z_k^4\bigr)\ ,$$ since $z_k\bar z_k=1$. When we sum this over $k$ from $1$ to $4n+2$ we only get a contribution from the constant term. Therefore we obtain $$S={1\over4}\left(\sum_{k=1}^{4n+2}\!{6\over16} \ -2\right)={6n-5\over16}\ .$$