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