Answer
$89,964$
Work Step by Step
... we want to apply $(8)\displaystyle \qquad \sum_{k=1}^{n}k^{3}=\left[\frac{n(n+1)}{2}\right]^{2}$,
but the index does not start at 1.
$\displaystyle \sum_{k=4}^{24}k^{3}=$ (terms from 4 to 24) = (24 terms) - (first 3 terms)
$=\displaystyle \sum_{k=1}^{24}k^{3}-\sum_{k=1}^{3}k^{3}$
... use formula (8)
$=\displaystyle \left[\frac{24(24+1)}{2}\right]^{2}-\left[\frac{3(3+1)}{2}\right]^{2}$
$=300^{2}-6^{2}$
$=89,964$
