simplify the summation of harmonic cosines

Question

Is it possible to simplify the following summation of harmonic cosines eliminating the sum?

\begin{equation} \sum_{n=1}^{N-1} \left( 1 + \cos \left( \frac{\pi n}{N} \right) \right)^N \end{equation}

The graph shows it will be like an exponential.

Answer 1

Let $\zeta_n=e^{i\pi/n}$; then $\zeta_n^{-k}(1+\zeta_n^k)^2=(1+\zeta_n^k)(1+\zeta_n^{-k})=2\big(1+\cos(k\pi/n)\big)$ and $$\sum_{k=1}^{2n}\zeta_n^{-kn}(1+\zeta_n^k)^{2n}=2^n\left(2^n+2\sum_{k=1}^{n-1}\Big(1+\cos\frac{k\pi}{n}\Big)^n\right).$$ On the other hand, by the binomial expansion, this sum equals $$\sum_{k=1}^{2n}\zeta_n^{-kn}\sum_{j=0}^{2n}\binom{2n}{j}\zeta_n^{kj}=\sum_{j=0}^{2n}\binom{2n}{j}\sum_{k=1}^{2n}\zeta_n^{(j-n)k}=2n\binom{2n}{n}$$ because $\sum\limits_{k=1}^{2n}\zeta_n^{mk}=0$ if $m$ is not an integer multiple of $2n$. Finally, $$\sum_{k=1}^{n-1}\Big(1+\cos\frac{k\pi}{n}\Big)^n=\frac{n}{2^n}\binom{2n}{n}-2^{n-1}.$$