Answer
See below.
Work Step by Step
1. Test for $n=1$, we have $LHS=1^2=1$ and $RHS=\frac{1}{6}(1)(1+1)(2(1)+1)=1$, thus it is true.
2. Assume the formula is true for $n=k$, we have
$1^2+2^2+3^2+...+k^2=\frac{1}{6}(k)(k+1)(2k+1)$
3. For $n=k+1$, we have $LHS=1^2+2^2+3^2+...+k^2+(k+1)^2=\frac{1}{6}(k)(k+1)(2k+1)+(k+1)^2=\frac{1}{6}(k+1)(2k^2+k+6k+6)=\frac{1}{6}(k+1)(2k^2+7k+6)=\frac{1}{6}(k+1)(k+2)(2k+3)=RHS$
4. We conclude that the formula is true for any $n$.
