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 $k=6$

Hint: think about (1+x)^n, and consult this section in DLMF. — Zau, Aug 19 '16 at 10:40
@SMA.D EXACTLY THIS IS MY QUESTION AND PEOPLE EDITED IT WRONG — satyajeet jha, Aug 19 '16 at 10:46
@SMA.D CAN YOU JUST EDIT THE QUESTION ? what u wrote is my question — satyajeet jha, Aug 19 '16 at 10:50
here k<=Nlook at the comment of @SMA.D .THIS IS WHAT MY QUESTION IS — satyajeet jha, Aug 19 '16 at 10:52
In your example k should be 2 then $2i\in{0,2,4}$ and $\sum_{i=0}^{k} \binom{N}{2i}= \binom{8}{0}+ \binom{8}{2}+\binom{8}{4}$ — SMA.D, Aug 19 '16 at 11:12
This is just the partial sum of $C(n-1,i)$ up to $i=2k$. There won't be anything that you would likely call a closed form, but there are asymptotics depending on how large $k$ is relative to $n$ (for instance, if $k \approx \alpha n$ for some fixed $0<\alpha<\tfrac12$ or if $k \approx \sqrt{n}$). Just search around for partial sum of binomial coefficients. — Erick Wong, Aug 19 '16 at 11:28

score 7 · Answer 1 · edited Apr 13 '17 at 12:58

7

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

edited Apr 13 '17 at 12:58

Community

1

answered Aug 19 '16 at 10:40

Robert Z

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1

your answer is just for $k=\lfloor{n/2}\rfloor$ the question wants $k\leq\lfloor{n/2}\rfloor$. – SMA.D Aug 19 '16 at 11:22
@SMA.D Thank you for pointing out. I misunderstood the question. – Robert Z Aug 19 '16 at 13:17

score 5 · Answer 2 · answered Aug 19 '16 at 10:43

5

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

answered Aug 19 '16 at 10:43

S.C.B.

22,768

the question has been edited properly – satyajeet jha Aug 19 '16 at 11:00

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

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