# Sum of Even Index Binomial Coefficients/Proof 2

## Theorem

$\displaystyle \sum_{i \mathop \ge 0} \binom n {2 i} = 2^{n - 1}$

## Proof

Let $N^*_n$ be the initial segment of natural numbers, one-based.

Let:

$E_n = \left\{ {X : \left({X \subset N^*_n}\right) \land \left({2 \mathrel \backslash \left\vert X \right\vert}\right)}\right\}$
$O_n = \left\{ {X : \left({X \subset N^*_n}\right) \land \left({2 \nmid \left\vert X \right\vert}\right)}\right\}$

That is:

$E_n$ is the set of all subsets of $N^*_n$ with an even number of elements
$O_n$ is the set of all subsets of $N^*_n$ with an odd number of elements.
$\displaystyle \sum_{i \mathop \ge 0} \binom n {2 i} = 2^{n - 1} \iff \left\vert{E_n}\right\vert = 2^{n - 1}$

which is to be proved by induction below.

### Basis of the Induction

Let $n = 1$.

Then

$\left\vert E_n \right\vert = \left\vert \left\{ {\varnothing}\right\} \right\vert = 1$

and:

$2^{n - 1} = 2^{1 - 1} = 2^0 = 1$

This is the basis for the induction.

### Induction Hypothesis

This is the induction hypothesis:

$\left\vert E_k \right\vert = 2^{k - 1}$

So:

$\left\vert O_k \right\vert = \left\vert 2^{N^*_k} \mathrel \backslash E_k \right\vert = 2^k - 2^{k-1} = 2^{k - 1}$

### Induction Step

This is the induction step:

 $\displaystyle \left\vert {E_{k + 1} }\right\vert$ $=$ $\displaystyle \left\vert{\left\{ {X : \left({X \subset N^*_{k + 1} }\right) \land \left({2 \mathrel \backslash \left\vert{X}\right\vert }\right)}\right\} }\right\vert$ Definition of $E_n$ $\displaystyle$ $=$ $\displaystyle \left\vert \left\{ {X \vert X \in E_k}\right\} \cup \left\{ {X \cup \left\{ {k+1}\right\} \vert X \in O_k}\right\} \right\vert$ Construction of the set with smaller sets, considering the presence of the element $k+1$ $\displaystyle$ $=$ $\displaystyle \left\vert{E_k}\right\vert + \left\vert{O_k}\right\vert$ $E_k$ and $O_k$ are disjoint by definition $\displaystyle$ $=$ $\displaystyle 2^{k - 1} + 2^{k - 1}$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle 2^k$

The result follows by induction.

$\blacksquare$