Sum of Even Index Binomial Coefficients

Theorem

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

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

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

Proof
From Sum of Binomial Coefficients for Given n we have:
 * $$\sum_{i \in \Z} \binom n i = 2^n$$

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

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

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

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

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

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

as the odd index coefficients cancel out.

Dividing by $$2$$ throughout gives us the result.