What's $\sum_{k=0}^n\binom{n}{2k}$? How do you calculate $\displaystyle \sum_{k=0}^n\binom{n}{2k}$? And doesn't the sum terminate when  2k exceeds n, so the upper bound should be less than n?
EDIT: I don't understand the negging. Have I violated a rule of conduct or something? 

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$\ds{\sum_{k = 0}^{n}{n \choose 2k}:\ {\large ?}}$

$$
\mbox{Note that}\quad\sum_{k = 0}^{n}{n \choose 2k}
=\sum_{k = 0}^{\color{#c00000}{\large\infty}}{n \choose 2k}.
$$

$$
\mbox{We'll use the identity}\quad
{m \choose s}=\oint_{0\ <\ \verts{z}\ =\ a}
{\pars{1 + z}^{m} \over z^{s + 1}}\,{\dd z \over 2\pi\ic}\,,\quad s \in {\mathbb N}
$$

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

