How to prove this equation? [closed]

Question

How to prove this equation: \begin{align}\sum_{k=1}^{n}\cos ^{2m+1} \frac{(2k-1)\pi}{2n+1} =\frac{1}{2}\end{align} where $k,m,n\in \mathbb{N}^*$.

Answer 1

The given sum is $S=\sum_{k=1}^{n} \cos(\frac{(2k-1)\pi}{2n+1})^{2m+1}$. Now, expanding the cosine in exponential $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$ we rewrite the sum as

$$S=\frac{1}{2^{2m+1}}\sum_{r=0}^{m} \binom{2m+1}{r} A(r,n)$$

here $$A(r,n)=\sum_{k=-n}^{n-1} e^{ia_r\frac{(2k+1)\pi}{2n+1}}$$

and $a_r=2m+1-2r$

Now, $A(r,n)=e^{-\frac{(2n-1)}{2n+1}i\pi a_r}\left(\frac{e^{i\frac{2n}{2n+1}2\pi a_r}-1}{e^{\frac{i2\pi a_r}{2n+1}}-1}\right)$

manipuating the terms in the exponential we get $A(r,n)=(-1)^{a_r}e^{\frac{2\pi a_r}{2n+1}}\left(\frac{e^{-\frac{i2\pi a_r}{2n+1}}-1}{e^{\frac{i2\pi a_r}{2n+1}}-1}\right)=(-1)^{a_r+1}=1$ (since $a_r=2m+1-2r$ is odd)

Hence, $S=\frac{1}{2^{2m+1}}\sum_{r=0}^{m} \binom{2m+1}{r}=\frac{1}{2}$