Sum of Even Index Binomial Coefficients/Proof 1

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
From Sum of Binomial Coefficients over Lower Index we have:
 * $\displaystyle \sum_{i \mathop \in \Z} \binom n i = 2^n$

That is:
 * $\displaystyle \binom n 0 + \binom n 1 + \binom n 2 + \binom n 3 + \cdots + \binom n n = 2^n$

as $\displaystyle \binom n i = 0$ for $i < 0$ and $i > n$.

This can be written more conveniently as:
 * $\displaystyle \binom n 0 + \binom n 1 + \binom n 2 + \binom n 3 + \binom n 4 + \cdots = 2^n$

Similarly, from Alternating Sum and Difference of Binomial Coefficients for Given n we have:
 * $\displaystyle \sum_{i \mathop \in \Z} \left({-1}\right)^i \binom n i = 0$

That is:
 * $\displaystyle \binom n 0 - \binom n 1 + \binom n 2 - \binom n 3 + \binom n 4 - \cdots = 0$

Adding them together, we get:
 * $\displaystyle 2 \binom n 0 + 2 \binom n 2 + 2 \binom n 4 + \cdots = 2^n$

as the odd index coefficients cancel out.

Dividing by $2$ throughout gives us the result.