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