Answer
See explanations.
Work Step by Step
Step 1. Prove the formula is true when $n=1$: $LHS=1^2=1$, $RHS=\frac{1(2)(3)}{6}=1$, thus it is true for $n=1$.
Step 2. Assume the formula is true when $n=k$: we have $1^2+2^2+3^2+...+k^2=\frac{k(k+1)(2k+1)}{6}$
Step 3. Prove it is true for $n=k+1$: $LHS=1^2+2^2+3^2+...+k^2+(k+1)^2=\frac{k(k+1)(2k+1)}{6}+(k+1)^2
=\frac{k(k+1)(2k+1)+6(k+1)^2}{6}=\frac{(k+1)[k(2k+1)+6(k+1)]}{6}=\frac{(k+1)(2k^2+7k+6)}{6}$
$RHS=\frac{(k+1)(k+2)[2(k+1)+1]}{6}=\frac{(k+1)(k+2)(2k+3)}{6}=\frac{(k+1)(2k^2+7k+6)}{6}$
Thus, we have $LHS=RHS$ which proves the formula is also true for $n=k+1$
Step 4. Conclusion: with mathematical induction, we have proved that the formula is true for all natural numbers n.
