The sum of the first n squares

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

Inductive step. Assume the formula for $n$. Then
$$\begin{aligned} \sum_{k=1}^{n+1}k^2&=\frac{n(n+1)(2n+1)}6+(n+1)^2 \\ &=\frac{(n+1)\bigl[n(2n+1)+6(n+1)\bigr]}6 \\ &=\frac{(n+1)(2n^2+7n+6)}6. \end{aligned}$$
Since $2n^2+7n+6=(n+2)(2n+3)$, this equals $\frac{(n+1)(n+2)(2n+3)}6$, the formula at rank $n+1$.

2.
$$\frac1{n^3}\cdot\frac{n(n+1)(2n+1)}6=\frac16\cdot\frac{n+1}n\cdot\frac{2n+1}n\longrightarrow\frac16\cdot1\cdot2=\frac13.$$

Remark. This is the Riemann sum for $\int_0^1x^2\,dx=\frac13$: the same result obtained from an algebraic formula.