Seicchi28 wrote: Let $n$ be a positive integer greater than $1$. Evaluate the following summation: \[ \sum_{k=0}^{n-1} \frac{1}{1 + 8 \sin^2 \left( \frac{k \pi}{n} \right)}. \] See here.

Seicchi28 wrote: Let $n$ be a positive integer greater than $1$. Evaluate the following summation: \[ \sum_{k=0}^{n-1} \frac{1}{1 + 8 \sin^2 \left( \frac{k \pi}{n} \right)}. \] Let $P_n(x)$ be the Chebychev polynomial of first kind such that $\cos nx=P_n(\cos x)$ Let $U_n(x)$ be the Chebychev polynomial of second kind such that $\sin (n+1)x=\sin xU_n(\cos x)$ Taking derivative of "$\cos nx=P_n(\cos x)$", we get $P'_n(x)=nU_{n-1}(x)$ Let then $Q_n(x)=P_n(\frac{5-x}4)-1$. $Q_n(x)$ has roots $\{1+8\sin^2\frac{k\pi}n\}_{k=0}^{n-1}$ So $S_n=\sum_{k=0}^{n-1}\frac 1{1+8\sin^2\frac{k\pi}n}=-\frac{Q'(0)}{Q(0)}=\frac 14\frac{P_n'(\frac 54)}{P_n(\frac 54)-1}$ $S_n=\frac n4\frac{U_{n-1}(\frac 54)}{P_n(\frac 54)-1}$ From $P_0(x)=1$, $P_1(x)=x$ and $P_{n+2}(x)=2xP_{n+1}(x)-P_n(x)$, we easily get $P_n(\frac 54)=2^{n-1}+2^{-n-1}$ And also $U_0(x)=1$, $U_1(x)=x$ and $U_{n+2}(x)=2xU_{n+1}(x)-U_n(x)$, we easily get $U_n(\frac 54)=\frac 13(2^{n+2}-2^{-n})$ And so $S_n=\frac n{12}\frac{2^{n+1}-2^{1-n}}{2^{n-1}+2^{-n-1}-1}=\frac n3\frac{2^{2n}-1}{2^{2n}-2^{n+1}+1}$ $=\frac n3\frac{(2^n-1)(2^n+1)}{(2^n-1)^2}$ And so $\boxed{\sum_{k=0}^{n-1}\frac 1{1+8\sin^2\frac{k\pi}n}=\frac n3\frac{2^n+1}{2^n-1}}$