Answer
$$\frac{1}{3}$$
Work Step by Step
$$\lim_{n\to\infty}\sum_{i=1}^n\frac{1}{n^3}\left(i-1\right)^2$$
$$\sum_{i=1}^n\frac{1}{n^3}\left(i-1\right)^2=\frac{1}{n^3}\sum_{i=1}^n(i-1)^2=\frac{1}{n^3}\sum_{i=1}^n(i^2-2i+1)$$
$$=\frac{1}{n^3}\left(\sum_{i=1}^ni^2-2\sum_{i=1}^ni+\sum_{i=1}^n1\right)$$
$$=\frac{1}{n^3}\left(\frac{n(n+1)(2n+1)}{6}-2\times\frac{n(n+1)}{2}+n\right)$$
$$=\frac{1}{n^3}\left(\frac{n(n+1)(2n+1)}{6}-n(n+1)+n\right)$$
$$=\frac{1}{n^3}\left(\frac{n(n+1)(2n+1)}{6}-n^2\right)$$
$$=\frac{n(n+1)(2n+1)}{6n^3}-\frac{n^2}{n^3}$$
$$=\frac{(n+1)(2n+1)}{6n^2}-\frac{1}{n}=\frac{2n^2+3n+1}{6n^2}-\frac{1}{n}$$
$$\lim_{n\to\infty}\frac{2n^2+3n+1}{6n^2}-\frac{1}{n}=\frac{2}{6}-0=\frac{1}{3}$$
