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 we have:
1) For $n=1: 1=1(3(1)-1)/2$.
2) Assume for $n=k: 1+4++3k-2=\frac{k(3k-1)}{2}$. Then for $n=k+1$:
$1+4++3k-2+3(k+1)-2=\frac{k(3k-1)}{2}+3(k+1)-2=\frac{k(3k-1)}{2}+3k+1=\frac{(k+1)(3k+2)}{2}=\frac{(k+1)(3(k+1)-1)}{2}.$
Thus we proved what we wanted to.