$\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.
 A: 
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 $j\mid n$
A: The first thing to note is that 2017 is prime so that for $k\ne 0$ or $2017$ the numbers $1,\omega^k,\omega^{2k},\dots,\omega^{2016k}$ are just a rearrangement of $1,\omega,\omega^2,\dots,\omega^{2016}$ and hence sum to 0.
For each $k$ we expand by the binomial theorem. The zero powers are each ${2017\choose 0}$ and so summed over $k$ they sum to $2017$. The $2017$th powers are each ${2017\choose 2017}$ and also sum to $2017$. But all the other powers sum to 0. So the grand total is just $4034$.
A: $$(1+\omega^k)^{2017}=\sum_{i=0}^{2017}\binom{2017}i\omega^{ki}\implies$$$${}$$
$$\sum_{k=0}^{2016}\sum_{i=0}^{2017}\binom{2017}i\omega^{ki}=\sum_{i=0}^{2017}\binom{2017}i\sum_{k=0}^{2016}\left(\omega^i\right)^k=\;\sum_{i=0}^{2017}\binom{2017}i\frac{1-\overbrace{\left(\omega^i\right)^{2017}}^{=1}}{1-\omega^i}=0$$
