Sum of Even Index Binomial Coefficients

Theorem

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

where $\displaystyle \binom n i$ is a binomial coefficient.

That is:
 * $\displaystyle \binom n 0 + \binom n 2 + \binom n 4 + \cdots = 2^{n-1}$

Proof 2
Let $N^*_n$ be the initial segment of natural numbers, one-based.

Let $E_n = \left\{ X \vert \left({X \subset N^*_n}\right) \wedge \left({2 \mid \left\vert X \right\vert}\right) \right\}$.

Let $O_n = \left\{ X \vert \left({X \subset N^*_n}\right) \wedge \left({2 \nmid \left\vert X \right\vert}\right) \right\}$.

So:
 * $\displaystyle \sum_{i \mathop \ge 0} \binom n {2i} = 2^{n-1} \Longleftrightarrow \left\vert E_n \right\vert = 2^{n-1}$

and the latter will be proved by induction below.

Base case
If $n=1$, then


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

and


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

So the base case holds.

Induction Hypothesis
Induction Hypothesis: $\left\vert E_k \right\vert = 2^{k-1}$.

So:


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

Induction Step
This is our Induction Step:

The result follows by induction.