I have to prove that $$\sum\_{i=0}^n {{n \choose i} \times 2^i} = 3^n $$ 
Such that ${n \choose i} = \frac{n!}{i!(n-i)!} $ and $n $ is some arbitrary int 
I proved we can expand 2^I in a way such that $$2^i = {i \choose 0} +{i \choose 1} + ... + {i \choose i} $$
But this doesn't seem to approach the proof is want so I tried to expand out ${n \choose i}$ and again this got ugly and led no where. I can't quite seem to find any approach that would lead to $3$ of any power.

I know this proof is true for the base case$$\sum\_{i=0}^0 {{0 \choose i} \times 2^i} = 1 => {0 \choose 0} \times 2^0 = 1 => 1 = 1 $$
Could anyone give me a hint to lead me in the right direction?

EDIT:
Wow, I always feel so stupid when I miss tiny identities such as that! Thank you everyone! I got my answer, I wish I could give all of you checks for the advice.