Answer
See below.
Work Step by Step
Proofs using mathematical induction consist 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 $n=1: 2=3^1-1$.
2) Assume for $n=k: 2+6+...2\cdot3^{k-1}=3^k-1$. Then for $n=k+1$:
$2+6+...2\cdot3^{k-1}+2\cdot3^k=3^k-1+2\cdot3^k=3\cdot3^k-1=3^{k+1}-1.$
Thus we proved what we wanted to.