Answer
\[
\frac{(-1+2 n)(-1+n)}{6}
\]
Work Step by Step
Remember that:
\[
\sum_{k=1}^{n} k^{2}=\frac{(1+n)(1+2 n)n}{6}
\]
We find:
$\sum_{k=1}^{n-1} \frac{k^{2}}{n}=
\frac{1}{n} \sum_{k=1}^{n-1} k^{2}=\frac{1}{n} \cdot \frac{(-1+1+n)((-1+n)2+1) (-1+n)}{6}=\frac{(-1+2 n)(-1+n)}{6}$
