$\displaystyle\sum_{k\geq 0}^{} \binom{n}{2k} = \sum_{k \geq 0}^{} \binom{n}{2k+1}$

Question

I read in twitter that: $$\displaystyle\sum_{k\geq 0}^{} \binom{n}{2k} = \sum_{k \geq 0}^{} \binom{n}{2k+1}$$ So far I've noticed that if $n$ is odd then $n=2k+1$ and we have that $$\binom{2k+1}{k-i} = \binom{2k+1}{k+1+i} $$ for $i= 0,1,\dots,k$

Since we must have that one of $k-i$ , $k+1+i$ must be odd and the other even for all $i=0,1,\dots,k$ then we must have that $$\displaystyle\sum_{l\geq 0}^{} \binom{2k+1}{2l} = \sum_{l \geq 0}^{} \binom{2k+1}{2l+1}$$ I don't know what to do when $n$ is even or if there's a way to prove this without breaking it into cases.

Hint: What is the binomial expansion of $(x+y)^n$? What is the binomial expansion of $(1+(-1))^n$? — JMoravitz, Feb 05 '20 at 15:05
Depending on your preferred style/flavor of proof and what makes the most sense to you, you would do well to go to the linked question and look at the questions that are linked to/from that page to see dozens of other times this question has come up and the many solutions here, such as this, this and others — JMoravitz, Feb 05 '20 at 15:14

score 3 · Accepted Answer · answered Feb 05 '20 at 15:05

3

A suggestion: Look at $(1-1)^n$.

answered Feb 05 '20 at 15:05

Christian Blatter

226,825

$\displaystyle\sum_{k\geq 0}^{} \binom{n}{2k} = \sum_{k \geq 0}^{} \binom{n}{2k+1}$

1 Answers1