# 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.

• Brute-force on the small N and find this sequence on the OEIS? – kotomord Nov 8 '16 at 12:23
• oeis.org/… - your sequence – kotomord Nov 8 '16 at 12:36

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.

• I came up with a slightly different way to approach this. But your way of visualizing is helpful as well. – Deepankar Arya Nov 8 '16 at 12:45

After posting this question, it struck me that

RHS = $4\sum {n \choose 4k} -2\sum {n \choose 4k} -2\sum {n \choose 4k + 2}) = 4\sum {n \choose 4k} - 2 \sum {n \choose 2k} = 4\sum {n \choose 4k} - 2 * 2^{n - 1} = 4\sum {n \choose 4k} - 2^{n} \implies \sum {n \choose 4k} = \frac{(1 + i)^n + (1 - i)^n + 2^n}{4}$

$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ \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 $$\ds{n = 0}$$.

• What is $\delta$ here? – YiFan Feb 7 at 23:10
• @YiFan $\displaystyle\texttt{Kronecker Delta}$: $\displaystyle \delta_{n0} = 1$ if $\displaystyle n = 0$ and $\displaystyle = 0$ otherwise. – Felix Marin Feb 8 at 23:13