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 $n=1: 1(3)=\frac{1(1+1)(2(1)+7)}{6}$.
2) Assume for $n=k: 1(3)+2(4)+k(k+2)=\frac{k(k+1)(2k+7)}{6}$. Then for $n=k+1$:
$1(3)+2(4)+k(k+2)+(k+1)(k+3)=\frac{k(k+1)(2k+7)}{6}+(k+1)(k+3)=\frac{k(k+1)(2k+7)}{6}+k^2+k+3k+3=\frac{(k+1)(k+2)(2k+9)}{6}=\frac{(k+1)(k+1+1)(2(k+1)+7)}{7}.$
Thus we proved what we wanted to.