Let $\omega=e^{i2\pi/2015}$, evaluate $\sum\limits_{k=1}^{2014}\frac{1}{1+\omega^k+\omega^{2k}}$

For simplicity I will write $n=2015$. Since $X^n-1=\prod\limits_{k=0}^{n-1}(X-\omega^k)$ we conclude that $$\frac{nX^{n-1}}{X^n-1}=\sum_{k=0}^{n-1}\frac{1}{X-\omega^k}$$ Substituting $X=j=e^{2i\pi/3}$ and $X= \overline{j}$ and then subtracting we get $$\frac{nj^{n-1}}{j^n-1}-\frac{n\bar{j}^{n-1}}{\bar{j}^n-1}=\sum_{k=0}^{n-1}\left(\frac{1}{j-\omega^k}-\frac{1}{\bar{j}-\omega^k}\right)=(\bar{j}-j)\sum_{k=0}^{n-1}\frac{1}{1+\omega^k+\omega^{2k}} $$ Finally, since $n=2015\equiv 2\mod 3$ we see that $j^{n-1}=j$ and $j^n=j^2=\bar{j}$, we get $$\frac{1}{-i\sqrt{3}}\left(\frac{nj}{j^2-1}-\frac{n\bar{j}}{j-1}\right)=\frac{1}{3}+\sum_{k=1}^{n-1}\frac{1}{1+\omega^k+\omega^{2k}}$$ The final step is easy and we get $$\sum_{k=1}^{n-1}\frac{1}{1+\omega^k+\omega^{2k}}=\frac{2n-1}{3}=1343$$ This conclusion is valid for every $n$ which is equal to $2\pmod3$.