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? 
 A: First of all, $\dbinom{n}{m} = 0$ if $m > n$. Hence, $\displaystyle\sum_{k = 0}^{n}\dbinom{n}{2k} = \sum_{k = 0}^{\lfloor n/2 \rfloor}\dbinom{n}{2k} = \sum_{\substack{0 \le m \le n \\ m \ \text{is even}}}\dbinom{n}{m}$.
To help you calculate the sum, note that by the binomial theorem: 
$2^n = (1+1)^n = \displaystyle\sum_{m = 0}^{n}\dbinom{n}{m}1^m = \sum_{\substack{0 \le m \le n \\ m \ \text{is even}}}\dbinom{n}{m} + \sum_{\substack{0 \le m \le n \\ m \ \text{is odd}}}\dbinom{n}{m}$
$0 = (1-1)^n = \displaystyle\sum_{m = 0}^{n}\dbinom{n}{m}(-1)^m = \sum_{\substack{0 \le m \le n \\ m \ \text{is even}}}\dbinom{n}{m} - \sum_{\substack{0 \le m \le n \\ m \ \text{is odd}}}\dbinom{n}{m}$. 
Do you see how to finish?
A: Recall that each binomial coefficient is the sum of the two above it in Pascal's triangle.
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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}

A: $$\sum_{k=1}^n \binom{n}{2k}=2^{n-1}-1$$
BECAUSE:
$$\sum_{k=1}^n \binom{n}{2k}=\binom{n}{2}+\binom{n}{4}+\binom{n}{3}+ \dots$$
$$(x+y)^n=\sum_{k=0}^{n} \binom{n}{k} x^k y^{n-k}$$
$$x=1,y=1: \sum_{k=0}^n  \binom{n}{k}=\binom{n}{1}+\binom{n}{2}+\binom{n}{3}+ \dots=2^n$$
$$\sum_{k=1}^n \binom{n}{2k}=\sum_{k=0}^n \binom{n}{2k}-1=\frac{\sum_{k=0}^n \binom{n}{k}}{2}-1=2^{n-1}-1$$
