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? 

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