Answer
See below.
Work Step by Step
Proofs using mathematical induction consists of two steps:
1) The base case: here we prove that the statement holds for the first natural number.
2) The inductive step: assume the statement is true for some arbitrary natural number $n$ larger than the first natural number, then we prove that then the statement also holds for $n + 1$.
Hence here: 1) For $n=1: 1=\frac{3^1-1}{2}$.
2) Assume for $n=k: 1+3+...+3^{k-1}=\frac{3^k-1}{2}$. Then for $n=k+1$: $1+3+...+3^{k-1}+3^k=\frac{3^k-1}{2}+3^k=\frac{3^k-1+2\cdot3^k}{2}=\frac{3\cdot3^k-1}{2}=\frac{3^{k+1}-1}{2}$
Thus we proved what we wanted to.