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!

Inductive Step: Assume that the expression holds for $n$. We want to prove that the expression holds also for $n+1$. That is $$\begin{align*}\sum_{i=0}^{n+1} 3^i &=\sum_{i=0}^n 3^i +3^{n+1}=^{\text{assumption of inductive step}}\\&= \frac {(3^{n+1})-1}{2}+3^{n+1}=\frac{3^{n+1}-1+2\cdot3^{n+1}}{2}=\frac{3\cdot3^{n+1}-1}{2}=\frac{3^{n+2}-1}{2}\\&=\frac{3^{(n+1)+1}-1}{2}\end{align*}$$ which proves that the expression holds also for $n+1$.
