$\sum_{k=0}^{2016}(1+ \omega^k)^{2017}\ $ Let $\omega  \ $be a root of the polynomial $\ x^{2016} +x^{2015}+x^{2014}+...+x+1=0  \ $. Then find the value of the following sum: $$\sum_{k=0}^{2016}(1+ \omega^k)^{2017}\ $$
Well I have simplified the expression as:
$$\sum_{k=0}^{2016}(1+ 2017\omega^k)\ $$ after that i further simplified it as 
$$\ 2016+2017\sum_{k=0}^{2016}(\omega^k)\ $$ but after this i am unable to proceed.


Let $\omega=\exp\left(\frac{2\pi i}{ n}\right)$ be the $n$-th root of unity. The following is valid
  \begin{align*}
\sum_{k=0}^{n-1}\left(1+\omega^k\right)^n=2n
\end{align*}
We obtain
  \begin{align*}
\sum_{k=0}^{n-1}\left(1+\omega^k\right)^n&=\sum_{k=0}^{n-1}\sum_{j=0}^{n}\binom{n}{j}\omega^{kj}\\
&=\sum_{j=0}^{n}\binom{n}{j}\sum_{k=0}^{n-1}\left(\omega^j\right)^k\\
&=2n+\sum_{j=1}^{n-1}\binom{n}{j}\sum_{k=0}^{n-1}\left(\omega^j\right)^k\tag{1}\\
&=2n+\sum_{{j=1}\atop{j\nmid n}}^{n-1}\binom{n}{j}\frac{1-\omega^{nj}}{1-\omega^j}\tag{2}\\
&=2n
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we separate the summands $j=0$ and $j=n$

*In (2) we use $\omega^n=1$ and have to respect that the sum is zero if $j\mid n$
