Answer
$\sum_{k=1}^{n-1} \frac{k^{3}}{n^{2}}=\frac{(-1+n)^{2}}{4}$
Work Step by Step
We find:
$\begin{aligned} \sum_{k=1}^{n-1} \frac{k^{3}}{n^{2}} &=\frac{1}{n^{2}} \sum_{k=1}^{n-1} k^{3} \\ &=\left[\frac{(-1+n)n}{2}\right]^{2}\frac{1}{n^{2}} \\ &=\frac{(n-1)^{2} n^{2}}{4 n^{2}} \\ &=\frac{(-1+n)^{2}}{4} \end{aligned}$
