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:
1) For $n=1: 1=\frac{1}{2}(3(1)-1)$
2) Assume for $n=k: 1+4+...+(3k-2)=\frac{k}{2}(3(k)-1)$. Then for $n=k+1:1+4+...+(3k-2)+(3k+1)=\frac{k}{2}(3(k)-1)+(3k+1)=1.5k^2-0.5k+3k+1=1.5k^2+2.5k+1=\frac{k+1}{2}(3k+2)=\frac{k+1}{2}(3(k+1)-1)$
Thus we proved what we wanted to.
