Sum of Even Index Binomial Coefficients/Proof 2

Proof
Let ${\N_n}^*$ be the initial segment of natural numbers, one-based.

Let:
 * $E_n = \set {X : \paren {X \subset {\N_n}^*} \land \paren {2 \divides \size X} }$
 * $O_n = \set {X : \paren {X \subset {\N_n}^*} \land \paren {2 \nmid \size X} }$

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:
 * $\ds \sum_{i \mathop \ge 0} \binom n {2 i} = 2^{n - 1} \iff \size {E_n} = 2^{n - 1}$

which is to be proved by induction below.

Basis of the Induction
Let $n = 1$.

Then


 * $\size {E_n} = \size {\set \O} = 1$

and:


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

This is the basis for the induction.

Induction Hypothesis
This is the induction hypothesis:


 * $\size {E_k} = 2^{k - 1}$

So:


 * $\size {O_k} = \size {2^{ {\N_k}^*} \divides E_k} = 2^k - 2^{k - 1} = 2^{k - 1}$

Induction Step
This is the induction step:

The result follows by induction.