Sum of Even Index Binomial Coefficients/Proof 2

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.

So by Cardinality of Set of Subsets:
 * $\displaystyle \sum_{i \mathop \ge 0} \binom n {2 i} = 2^{n - 1} \iff \left\vert{E_n}\right\vert = 2^{n - 1}$

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:

The result follows by induction.