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 $n=1: 1=1(2(1)-1)=1$.
2) Assume for $n=k: 1+5+...+(4k-3)=k(2k-1)$. Then for $n=k+1$:
$1+5+...+(4k-3)+(4k+1)=k(2k-1)+(4k+1)=2k^2-k+4k+1=(k+1)(2k+1)=(k+1)(2(k+1)-1).$
Thus we proved what we wanted to.