Proving $\sum_{k=1}^{n}\frac{1-a\cos\left(\frac{2\pi k}{n}\right)}{1-2a\cos\left(\frac{2\pi k}{n}\right)+a^{2}}=\frac{n}{1-a^{n}}$

Question

How do prove this relation? $$\sum_{k=1}^{n}\frac{1-a\cos\left(\frac{2\pi k}{n}\right)}{1-2a\cos\left(\frac{2\pi k}{n}\right)+a^{2}}=\frac{n}{1-a^{n}} \text{ } \text{ } \text{ } \text{ } \text{ }\text{ }\text{ }\text{ }\text{ }\text{ } a\in(-1,1)$$

Answer 1

let's start with a function like this: $$f\left(x\right)=x^n-1\Rightarrow f'\left(x\right)=nx^{n-1}$$ $$x^n-1=0\Rightarrow \left[x_k=e^{\frac{2\pi ki}{n}}\right]_{k=1,2,3,..n}$$ From partial fraction we have: $$\frac{1}{f\left(x\right)}=\frac{1}{x^n-1}=\sum _{k=1}^n\frac{A_k}{x-x_k}$$

$$\boxed{A_k=\frac{1}{f'\left(x_k\right)}=\frac{1}{n\left(e^{\frac{2\pi ik}{n}}\right)^{n-1}}=\frac{e^{\frac{2\pi ik}{n}}}{n}}$$

$$\frac{n}{x^n-1}=\sum _{k=1}^n\frac{e^{\frac{2\pi ik}{n}}}{x-e^{\frac{2\pi ik}{n}}}$$ $$\frac{n}{x^n-1}=\sum _{k=1}^n\frac{1}{e^{-\frac{2\pi ik}{n}}x-1}$$ $$\frac{n}{x^n-1}=\sum _{k=1}^n\frac{1}{x\cos \left(\frac{2\pi k}{n}\right)-xi\sin \left(\frac{2\pi k}{n}\right)-1}$$ Now we take real part from both side: $$\boxed{\Re \frac{1}{a+bi}=\frac{a}{a^2+b^2}}$$

$$\frac{n}{x^n-1}=\sum _{k=1}^n\frac{x\cos \left(\frac{2\pi k}{n}\right)-1}{x^2-2x\cos \left(\frac{2\pi k}{n}\right)+1}$$ So the proof is done.