Prove by induction:$\sum_{i=0}^n 3^i = \frac {(3^{n+1})-1}{2}$ Prove by induction: $$\sum_{i=0}^n 3^i = \frac {(3^{n+1})-1}{2}$$
Basis: For $n=0$ we have $1 = 1$
Inductive Step: Now this is where I don't know what to do, any kind of help would be much appreciated. Thanks!

You have the base step.
Induction hypothesis: Assume that $$\sum_{i=0}^k3^i=\frac{3^{k+1}-1}{2}$$
for some $k\ge0$
Inductive step:
Then $$\begin{align}\sum_{i=0}^{k+1}&=3^{k+1}+\sum_{i=0}^k3^i\\&=3^{k+1}+\frac{3^{k+1}-1}{2}\\&=\frac{3^{k+2}-1}{2}\end{align}$$
Thus our formula is proved.
