Answer
See explanations.
Work Step by Step
1. Confirm $S_1$ is true: for $n=1$, $LHS=1$, $RHS=(1)(3-2)=1=LHS$
2. Assume $S_k$ is true, we have $1+7+13+...+(6k-5)=k(3k-2)$
3. For $n=k+1$, we have $LHS=1+7+13+...+(6k-5)+(6(k+1)-5)=k(3k-2)+(6k+1)=3k^2+4k+1$ while $RHS=(k+1)(3(k+1)-2)=(k+1)(3k+1)=3k^2+4k+1=LHS$, thus $S_{k+1}$ is also true.
4. We have proved that $S_n$ is true for every positive integer $n$ through mathematical induction.
