TITLE: How to find sums like $\sum_{k=0}^{39} \binom{200}{5k}$
QUESTION [14 upvotes]: How do I find sums like these?--
$$S=\displaystyle\sum_{k=0}^{39} \dbinom{200}{5k}$$
that is, when there is a summation of binomial coefficients, but with jumps of some terms..?

REPLY [5 votes]: $\newcommand{\+}{^{\dagger}}
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$\ds{S\equiv\sum_{k = 0}^{39}{200 \choose 5k}:\ {\large ?}}$

$$
\mbox{Note that}\quad S = -1+
\color{#c00000}{\sum_{k = 0}^{\infty}{200 \choose 5k}}\tag{1}
$$

\begin{align}
&\color{#c00000}{\sum_{k = 0}^{\infty}{200 \choose 5k}}
=\sum_{k = 0}^{\infty}\oint_{\verts{z}\ =\ a\ >\ 1}
{\pars{1 + z}^{200} \over z^{5k + 1}}\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ a\ >\ 1}{\pars{1 + z}^{200} \over z}
\sum_{k = 0}^{\infty}\pars{1 \over z^{5}}^{k}
\,{\dd z \over 2\pi\ic}
\\[5mm]&=\oint_{\verts{z}\ =\ a\ >\ 1}{\pars{1 + z}^{200} \over z}
\,{1 \over 1 - 1/z^{5}}\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ a\ >\ 1}{z^{4}\pars{1 + z}^{200} \over z^{5} - 1}
\,{\dd z \over 2\pi\ic}
\end{align}

The integrand simple poles are the five roots of $\ds{z^{5} - 1 = 0}$. Namely,
  $$
z_{n} = \expo{2n\pi\ic/5}\,,\qquad\qquad n=-2,-1,0,1,2.
$$
  Then,

\begin{align}
&\color{#c00000}{\sum_{k = 0}^{\infty}{200 \choose 5k}}
=\sum_{n = -2}^{2}\lim_{z \to z_{n}}
\bracks{\pars{z - z_{n}}\,{z^{4}\pars{1 + z}^{200} \over z^{5} - 1}}
={1 \over 5}\sum_{n = -2}^{2}\pars{1 + z_{n}}^{200}
\\[5mm]&={1 \over 5}\sum_{n = -2}^{2}\pars{1 + \expo{2n\pi\ic/5}}^{200}
={1 \over 5}\sum_{n = -2}^{2}\pars{\expo{-n\pi\ic/5} + \expo{n\pi\ic/5}}^{200}
\\[5mm]&={2^{200} \over 5}\sum_{n = -2}^{2}\cos^{200}\pars{n\,{\pi \over 5}}
={2^{200} \over 5}\bracks{1 + 2\cos^{200}\pars{\pi \over 5}
+2\cos^{200}\pars{2\pi \over 5}}
\end{align}

Replacing this result in $\pars{1}$:

\begin{align}
S&\equiv\color{#66f}{\large\sum_{k = 0}^{39}{200 \choose 5k}}\
{\Large =}\
\color{#66f}{\large{2^{200} \over 5} - 1 +
{2^{201} \over 5}\,\bracks{\cos^{200}\pars{\pi \over 5}
+\cos^{200}\pars{2\pi \over 5}}}
\\[5mm]&\approx {\tt 3.2139 \times 10^{59}}
\end{align}