Evaluate $\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+...+\binom{n}{2k}$

Question

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 $k=6$

Answer 1

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 $n\geq 1$, $$\sum_{i \text{ even}}\binom{n}{i}=\sum_{i \text{ odd}}\binom{n}{i}=2^{n-1}.$$

Answer 2

By the binomial theorem,

$$ 0=(1-1)^n=\sum_{i=0}^n \binom{n}{i}(-1)^i=\sum_{i \equiv 0 \pmod 2 }\binom{n}{i}-\sum_{i \equiv 1 \pmod 2}\binom{n}{i}. $$

$$ 2^n=(1+1)^n=\sum_{i=0}^n \binom{n}{i}1^{i}=\sum_{i \equiv 0 \pmod 2 }\binom{n}{i}+\sum_{i \equiv 1 \pmod 2}\binom{n}{i}. $$

Thus $$\sum_{i \equiv 0 \pmod 2 }\binom{n}{i}=2^{n-1}$$