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 $n=1: 1=0.5(1)(3(1)-1)$.
2) Assume for $n=k: 1+4+...+(3k-2)=0.5k(3k-1)$. Then for $n=k+1$:
$1+4+...+(3k-2)+(3k+1)=0.5k(3k-1)+(3k+1)=1.5k^2-0.5k+3k+1=0.5(k+1)(3k+2)=0.5(k+1)(3(k+1)-1).$
Thus we proved what we wanted to.