Sum of binomial coefficients with index divisible by 4 I am trying to obtain a closed form expression for 
$\sum {n \choose 4k}$
I am trying to use the binomial expansion of $(1 + i)^n$ and $(1 - i)^n$.
$$(1 + i)^n + (1 - i)^n = 2\left(\sum {n \choose 4k} -\sum {n \choose 4k + 2}\right)$$ Stuck at this now. I can't come up with a way to simplify the second term on RHS. Any help would be appreciated.
 A: We have the followings : 
$$(1+1)^n=\binom n0+\binom n1+\binom n2+\binom n3+\binom n4+\binom n5+\cdots$$
$$(1-1)^n=\binom n0-\binom n1+\binom n2-\binom n3+\binom n4-\binom n5+\cdots$$
$$(1+i)^n=\binom n0+\binom n1i-\binom n2-\binom n3i+\binom n4+\binom n5i-\cdots$$
$$(1-i)^n=\binom n0-\binom n1i-\binom n2+\binom n3i+\binom n4-\binom n5i-\cdots$$
Adding these gives
$$(1+1)^n+(1-1)^n+(1+i)^n+(1-i)^n=4\left(\binom n0+\binom n4+\binom n8+\cdots\right)$$
I think that you can continue from here.
A: After posting this question, it struck me that 
RHS =  $4\sum {n \choose 4k} -2\sum {n \choose 4k} -2\sum {n \choose 4k + 2}) = 4\sum {n \choose 4k} - 2 \sum {n \choose 2k} = 4\sum {n \choose 4k} - 2 * 2^{n - 1} = 4\sum {n \choose 4k} - 2^{n} \implies \sum {n \choose 4k} = \frac{(1 + i)^n + (1 - i)^n + 2^n}{4}$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \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}}}
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\begin{align}
\left.\sum_{k = 0}^{\infty}{n \choose 4k}
\,\right\vert_{\ n\ \in\ \mathbb{N}_{\large n\ \geq\ 0}} & =
\sum_{k = 0}^{\infty}{n \choose k}{1^{k} + \pars{-1}^{k} + \ic^{k} + \pars{-\ic}^{k} \over 4}
\\[5mm] & =
{1 \over 4}\sum_{k = 0}^{\infty}{n \choose k}\pars{1}^{k} +
{1 \over 4}\sum_{k = 0}^{\infty}{n \choose k}\pars{-1}^{k} +
{1 \over 2}\,\Re\sum_{k = 0}^{\infty}{n \choose k}\ic^{k}
\\[5mm] & =
{1 \over 4}\pars{1 + 1}^{n} + {1 \over 4}\bracks{1 + \pars{-1}}^{n} +
{1 \over 2}\,\Re\pars{1 + \ic}^{n}
\\[5mm] & =
2^{n - 2} + {1 \over 4}\,\delta_{n0} + 
{1 \over 2}\,\Re\bracks{\root{1^{2} + 1^{2}}\expo{\ic\arctan\pars{1/1}}}^{n}
\\[5mm] & =
2^{n - 2} + {1 \over 4}\,\delta_{n0} +
{1 \over 2}\,\Re\bracks{2^{1/2}\expo{\ic\pi/4}}^{n}
\\[5mm] & =
\bbx{2^{n - 2} + {1 \over 4}\,\delta_{n0} +
2^{n/2 - 1}\cos\pars{n\pi \over 4}}
\end{align}

Note tat it yields the correct value when $\ds{n = 0}$.

