Answer
See below.
Work Step by Step
1. For $n=1$, we have $LHS=1^2=1$ and $RHS=\frac{1(1+1)(2(1)+1)}{6}=1=LHS$, thus P(1) is true.
2. Assume P(k) ($k\gt1$) is true, we have:
$1^2+2^2+...+k^2=\frac{k(k+1)(2k+1)}{6}$
3. For $n=k+1$, we have:
$LHS=1^2+2^2+...+k^2+(k+1)^{2}=\frac{k(k+1)(2k+1)}{6}+(k+1)^{2}\\
=\frac{k(k+1)(2k+1)+6(k+1)^2}{6}=\frac{(k+1)(2k^2+k+6k+6)}{6}=\frac{(k+1)(k+2)(2k+3)}{6}=RHS$
4. Thus $P(k+1)$ is also true and we have proved the statement by mathematical induction.