# Proving $\sum_{r=0}^{n-1} (-1)^r \cos^n{\left(\frac{r\pi}{n}\right)}=\frac{n}{2^{n-1}}$

Prove that $$\sum_{r=0}^{n-1} (-1)^r \cos^n{\left(\dfrac{r\pi}{n}\right)}=\dfrac{n}{2^{n-1}}$$

I proved the result using induction, however am more interested in finding the sum using complex numbers. Any hint?

The statement above seems to be erroneous. It can be shown that

$$S=\sum_{r=0}^{n-1}(-1)^r\cos^n(\frac{r\pi}{n})=\frac{n}{2^{n-1}}$$

On the LHS, rewrite $$\cos x=\frac{e^{ix}+e^{-ix}}{2}$$ and expand using the binomial theorem to get

$$S=\frac{1}{2^n}\sum_{k=0}^n{n\choose k}\sum_{r=0}^{n-1}(-1)^re^{\frac{i\pi r}{n}(2k-n)}$$

However,

$$\sum_{r=0}^{n-1}(-1)^re^{\frac{i\pi r}{n}(2k-n)}=\sum_{r=0}^{n-1}(e^{i2k\pi/n})^r=\frac{(e^{i2k\pi/n})^n-1}{e^{i2k\pi/n}-1}=n\sum_{m=-\infty}^{\infty}\delta_{k,mn}$$

and thus

$$S=\frac{n}{2^n}\Big({n\choose 0}+{n\choose n}\Big)=\frac{n}{2^{n-1}}$$

• How did you write $n\sum_{m=-\infty}^{\infty}\delta_{k,mn}$? Apr 24, 2020 at 19:05
• Denote $\omega=e^{\frac{i2k\pi}{n}}$, $\omega^n=1$. Then $\frac{\omega^n-1}{\omega-1}=0$ if $\omega\neq 1$ but $\lim_{\omega \to 1}\frac{\omega^n-1}{\omega-1}=n$, and that's why that equality is true, it just says that it's true for all $k=mn, m\in \mathbb{N}$, since those are the values for which $\omega=1$. Apr 24, 2020 at 19:55
• Thank you. That was gr8 Apr 24, 2020 at 20:29