How do you find the sum of Sigma (k+1)^2(k-3) where k is [2,5]?

${\sum}_{k = 2}^{5} {\left(k + 1\right)}^{2} \left(k - 3\right)$
With so few terms, since we're only adding $4$, it's easy to just plug in all $k$ values from $k = 2$ to $k = 5$ and add them:
$= {\left(2 + 1\right)}^{2} \left(2 - 3\right) + {\left(3 + 1\right)}^{2} \left(3 - 3\right) + {\left(4 + 1\right)}^{2} \left(4 - 3\right) + {\left(5 + 1\right)}^{2} \left(5 - 3\right)$
$= 9 \left(- 1\right) + 16 \left(0\right) + 25 \left(1\right) + 36 \left(2\right)$
$= \textcolor{b l u e}{88}$