Answer
$$
214365
$$
Work Step by Step
Remember that:
$\sum_{k=1}^{n} k^{3}=\left[\frac{(1+n)n}{2}\right]^{2}$
and $\sum_{k=1}^{n} k=\frac{(1+n)n}{2}$
We find:
$\sum_{k=1}^{30} (-2+k)(2+k)k=\sum_{k=1}^{30} -4 k+k^{3}=\sum_{k=1}^{30} -4+k^{3}\sum_{k=1}^{30} k=\left[\frac{30 \cdot(1+30)}{2}\right]^{2}-4\left[\frac{30 \cdot(1+30)}{2}\right]=214365$
