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?
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this communityI 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?
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.