Answer
$$
2856
$$
Work Step by Step
Remember that:
$$
\sum_{k=1}^{n} k^{2}=\frac{(1+n)(1+2 n)n}{6}
$$
We find:
$\sum_{k=4}^{20} k^{2}=\sum_{k=1}^{20} k^{2}-\sum_{k=1}^{3} k^{2}=\frac{20 \cdot(1+20) \cdot(2 \cdot 1+20)}{6}-\frac{3 \cdot(1+3) \cdot(2 \cdot 1+3)}{6}=-14+2870=2856$
