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

1 Answer

We show equality by showing the difference of both sums is zero.

We obtain \begin{align*} \sum_{{i=0}\atop{i \text{ even}}}^n&\binom{n}{i}-\sum_{{i=0}\atop{i \text{ odd}}}^n\binom{n}{i}\\\ &=\sum_{{i=0}\atop{i \text{ even}}}^n\binom{n}{i}(-1)^i+\sum_{{i=0}\atop{i \text{ odd}}}^n\binom{n}{i}(-1)^i\tag{1}\\ &=\sum_{i=0}^n\binom{n}{i}(-1)^i\tag{2}\\ &=(1-1)^n\tag{3}\\ &=0 \end{align*} and the claim follows.

Comment:

• 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.

• In (2) we simplify the expression by adding both sums.

• In (3) we apply the binomial theorem.