\sum_{k=0, k even}^n {n \choose k}*2^k = \frac{(3^n)+((-1)^n)}{2}

Question

Prove the following combinatoric identity:

$$\sum_{k=0, k even}^n {n \choose k}*2^k = \frac{(3^n)+((-1)^n)}{2}$$

Answer 1

Hint:

$\left(1+2\right)^{n}+\left(1-2\right)^{n}=\sum_{k}\binom{n}{k}2^{^{k}}+\sum_{k}\binom{n}{k}\left(-2\right){}^{k}$

Which terms fall away against eachother?