How to compute $\displaystyle\sum_{k\equiv 1\!\!\pmod{\!4}}\!\!\binom{2014}{k}$?

Question

I have to compute:

$S=\binom{2014}{1}+\binom{2014}{5}+\binom{2014}{9}+...+\binom{2014}{2009}+\binom{2014}{2013}$

Could someone help me ?

Answer 1

We have: $$\begin{eqnarray*}\sum_{k\equiv 1\!\!\pmod{\!4}}\binom{2014}{k}&=&\frac{1}{4}\sum_{k=0}^{2014}\binom{2014}{k}\left(i^{k-1}+(-1)^{k-1}+(-i)^{k-1}+1^{k-1}\right)\\&=&\frac{1}{4}\left(-i(1+i)^{2014}+i(1-i)^{2014}+2^{2014}\right)\\&=&\frac{1}{4}\left(i^2\cdot2^{1007}+i^2\cdot 2^{1007}+2^{2014}\right)\\&=&\frac{2^{2014}-2^{1008}}{4}\\&=&\color{red}{2^{2012}-2^{1006}}.\end{eqnarray*}$$