Prove that $\sum_{i=0}^n4^i$ = $1/3(4^{n+1} - 1)$ $\sum_{i=0}^n4^i$ = $1/3(4^{n+1} - 1)$
Attempt:
Let $n =0$ $4^0 = 1 \text{ and } (4^{0+1} -1)/3 = 1$
Assume true at $n = k \text{ so we have} \sum_{i=0}^k4^i = 1/3(4^{k+1} -1)$
The part I'm stuck at is the 3rd step. Can someone point me in the right direction after this?
$\sum_{i=0}^{k+1}4^i$ ?

Hint: $\displaystyle \sum_{i=0}^{k+1} 4^i = \displaystyle \sum_{i=0}^k 4^i + 4^{k+1}$
