Evaluate $\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+...+\binom{n}{2k}$ What is the closed formula for finding 

$$
\sum_{i=0}^{k} \binom{n}{2i}=?\quad ,k\leq \lfloor{n/2}\rfloor
$$

I need summation only up to some intermediate  $k$ where $k\leq \lfloor{n/2}\rfloor$ .
SAY I need a closed formula to calculate $\binom{8}{0}+ \binom{8}{2}+\binom{8}{4} +\binom{8}{6}$
HERE $n=8$ and $k=6$

We show a closed formula for $k=\lfloor n/2\rfloor$. For the general case, it is possible to give some asymptotic bounds. See here: https://mathoverflow.net/questions/17202/sum-of-the-first-k-binomial-coefficients-for-fixed-n
We have that: 
$$0=(1-1)^n=\sum_{i=0}^n \binom{n}{i}(-1)^i=\sum_{i \text{ even}}\binom{n}{i}-\sum_{i \text{ odd}}\binom{n}{i},$$
and
$$2^n=(1+1)^n=\sum_{i=0}^n \binom{n}{i}=\sum_{i \text{ even}}\binom{n}{i}+\sum_{i \text{ odd}}\binom{n}{i}.$$
Therefore for $n\geq 1$,
$$\sum_{i \text{ even}}\binom{n}{i}=\sum_{i \text{ odd}}\binom{n}{i}=2^{n-1}.$$
