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 the statement also holds for $n + 1$.
Hence here:
1) For $n=1: 3=3(1)/2(1+1)=3$.
2) Assume for $n=k: 3+6+9+...+3k=3k/2\cdot (k+1)$. Then for $n=k+1$:
$3+6+9+...+3k+3k+3=3k/2\cdot (k+1)+3k+3=3k^2/2+3k/2+3k+3=3(k+1)/2(k+2)=3(k+1)/2((k+1)+1).$
Thus we proved what we wanted to.