TITLE: Find the sum of $\binom{2016}{4} + \binom{2016}{8} +\binom{2016}{12} + \dots + \binom{2016}{2016}$
QUESTION [10 upvotes]: Problem:
Find 
$$\dbinom{2016}{4} + \dbinom{2016}{8} +\dbinom{2016}{12} + \dots + \dbinom{2016}{2016}$$
I don't know how to attempt this problem, other than that this sum is equivalent to finding the sum of the coefficients of degree 4 terms in the polynomial,
$$P(x) = (x+1)^{2016}$$
**I know that there's a duplicate of this problem somewhere, but I just can't find it on the website. Any help is appreciated!

REPLY [5 votes]: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{n = 1}^{504}{2016 \choose 4n} & =
-1 + \sum_{n = 0}^{2016}{2016 \choose n}
{1 + \pars{-1}^{n} + \ic^{n} + \pars{-\ic}^{n} \over 4}
\\[5mm] & =
-1 + {1 \over 4}\sum_{n = 0}^{2016}{2016 \choose n} +
{1 \over 4}\sum_{n = 0}^{2016}{2016 \choose n}\pars{-1}^{n}
\\[2mm] & +
{1 \over 2}\,\Re\sum_{n = 0}^{2016}{2016 \choose n}\ic^{n}
\\[5mm] & =
-1 + {1 \over 4}\,\pars{1 + 1}^{2016}+ {1 \over 4}\,\pars{1 - 1}^{2016} +
{1 \over 2}\,\Re\pars{1 + \ic}^{2016}
\\[5mm] & =
-1 + 2^{2014} + {1 \over 2}\,\Re\pars{2^{1008}\expo{504\pi\ic}} =
\bbx{\ds{-1 + 2^{2014} + 2^{1007}}}
\end{align}