TITLE: Finding $\binom{n}{0} + \binom{n}{3} + \binom{n}{6} + \ldots $
QUESTION [13 upvotes]: Help me to simplify:$$\binom{n}{0} + \binom{n}{3} + \binom{n}{6} + \ldots $$
I got a hunch that it will depend on whether $n$ is a multiple of $6$ and equals to $\frac{2^n+2}{3}$ when $n$ is a multiple of $6$.

REPLY [2 votes]: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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We can sum 'up to' $\ds{\infty\ \pars{~\mbox{Why ?}~}}$:

\begin{align}&\color{#66f}{\large\sum_{k = 0}^{\infty}{n \choose 3k}}
=\sum_{k = 0}^{\infty}\oint_{\verts{z}\ =\ a\ >\ 1}
{\pars{1 + z}^{n} \over z^{3k + 1}}\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ a\ >\ 1}
{\pars{1 + z}^{n} \over z}\sum_{k = 0}^{\infty}\pars{1 \over z^{3}}^{k}
\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\verts{z}\ =\ a\ >\ 1}
{\pars{1 + z}^{n} \over z}{1 \over 1 - 1/z^{3}}\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ a\ >\ 1}
{\pars{1 + z}^{n}z^{2} \over z^{3} - 1}\,{\dd z \over 2\pi\ic}
\end{align}

$\ds{z^{3} - 1 = 0}$ has roots at $\ds{z_{m} = \expo{2m\pi\ic/3}\,,\ m = -1,0,1}$:
\begin{align}&\color{#66f}{\large\sum_{k = 0}^{\infty}{n \choose 3k}}
=\sum_{m = -1}^{1}\lim_{z\ \to\ z_{m}}
\bracks{\pars{z - z_{m}}\,{\pars{1 + z}^{n}z^{2} \over z^{3} - 1}}
=\sum_{m = -1}^{1}\bracks{{\pars{1 + z_{m}}^{n}z_{m}^{2} \over 3z_{m}^{2}}}
\\[3mm]&={1 \over 3}\sum_{m = -1}^{1}\pars{1 + z_{m}}^{n}
={1 \over 3}\braces{2^{n} + 2\,\Re\pars{\bracks{1 + \expo{2\pi\ic/3}}^{n}}}
\\[3mm]&={1 \over 3}\bracks{%
2^{n} + 2\,\Re\pars{\expo{-n\pi\ic/3}2^{n}\cos^{n}\pars{\pi \over 3}}}
=\color{#66f}{\large{1 \over 3}\bracks{2^{n} + 2\cos\pars{n\pi \over 3}}}
\end{align}