Sum of squares by induction

Base case. $n=1$: $1=\frac{1\cdot2\cdot3}{6}=1$. ✓

Inductive step. Assume the formula for $n$. Then
$$\sum_{k=1}^{n+1}k^2=\frac{n(n+1)(2n+1)}{6}+(n+1)^2.$$
Factor $(n+1)$:
$$=\frac{(n+1)[n(2n+1)+6(n+1)]}{6}=\frac{(n+1)(2n^2+n+6n+6)}{6}=\frac{(n+1)(2n^2+7n+6)}{6}.$$
Factor $2n^2+7n+6=(n+2)(2n+3)$:
$$=\frac{(n+1)(n+2)(2n+3)}{6}.$$
This is the formula for $n+1$.

Conclusion. By induction, the formula holds.