#### 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 $n=1: 3=\frac{3(1)(1+1)}{2}$.
2) Assume for $n=k: 3+6+...+3k=\frac{3(k)(k+1)}{2}$. Then for $n=k+1$:
$3+6+...+3k+3k+3=\frac{3(k)(k+1)}{2}+3k+3=\frac{3k^2+3k}{2}+\frac{6k+6}{2}=\frac{3k^2+9k+6}{2}=\frac{3(k+1)(k+2)}{2}=\frac{3(k+1)((k+1)+1)}{2}.$
Thus we proved what we wanted to.