Combinatorially prove that $\sum_{i=0}^n {n \choose i} 2^i = 3^n $ So I'm not sure at all how to prove things using a combinatorial proof. Where to do I start? What do i need to think about etc.  For example how would i prove
$$\sum_{i=0}^n {n \choose i} 2^i = 3^n $$ 

Late answer but I thought a binomial theorem approach was pretty neat too. I think someone in the comments mentioned it. I also know op asked for a combinatorial proof but I noticed that after writing this so I will leave it anyway as a point of interest.
Recall the binomial theorem:
$$
(x+y)^n = \sum_{i=0}^n{n \choose i}x^{n-i}y^i
$$
To get the rhs to match your question, we can let $x=1$ and $y = 2$ giving us:
$$
\begin{align}
(1+2)^n &= \sum_{i=0}^n{n \choose i}1^{n-i}2^i \\
        &= \sum_{i=0}^n{n \choose i}2^i 
\end{align}
$$
So we have 
$$
\sum_{i=0}^n{n \choose i}2^i  = (1+2)^n = 3^n
$$
