How to prove that $\sum_{i-is-even}^n \binom{n}{i} = \sum_{i-is-odd}^n \binom{n}{i}$

I need to prove that

$\sum_{i-is-even}^n \binom{n}{i} = \sum_{i-is-odd}^n \binom{n}{i}$

i starts from 0.

I succeeded proving this for odd n. But how to prove it for even n's?

• Hint: Look at the binomial expansion of $(1+x)^n$ and consider $x=-1$ – user12345 Apr 20 '17 at 16:48
• Hint: Expand $\binom ni = \binom{n-1}{i-1} + \binom{n-1}i$. – Arthur Apr 20 '17 at 16:49
• Your assertion can be rephrased as: $\sum_{k=0}^{n} (-1)^{k} \binom{n}{k} = 0$. See math.stackexchange.com/q/94514/215011 – grand_chat Apr 20 '17 at 17:16
• If $n > 0$ !!!. – Felix Marin Apr 24 '17 at 2:10

• In (1) we multiply the terms with $(-1)^i$, which is $1$ if $i$ is even and $-1$ if $i$ is odd. This is just a preparation to easily merge both sums.