Sum of squares by induction

Base case. For $n=1$, the left side is $1$ and the right side is $\frac{1\cdot2\cdot3}{6}=1$.

Induction step. Assume the formula holds for some $n\ge1$. Then
$$\sum_{k=1}^{n+1}k^2=\frac{n(n+1)(2n+1)}{6}+(n+1)^2=\frac{(n+1)\left[n(2n+1)+6(n+1)\right]}{6}.$$
Since $n(2n+1)+6(n+1)=2n^2+7n+6=(n+2)(2n+3)$, this equals $\dfrac{(n+1)(n+2)(2n+3)}{6}$, which is the formula for $n+1$.

By induction, the formula holds for all $n\ge1$.