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 $n=1: 3=\frac{1(1+5)}{2}$.
2) Assume for $n=k: 3+4+...+k+2=\frac{k(k+5)}{2}$. Then for $n=k+1$:
$3+4+...+k+2+k+3=\frac{k(k+5)}{2}+k+3=\frac{k^2+5k}{2}+\frac{2k+6}{2}=\frac{k^2+7k+6}{2}=\frac{(k+1)(k+6)}{2}=\frac{(k+1)((k+5)+1)}{2}.$
Thus we proved what we wanted to.