Problem

(a) Use the standard results for summations to show that, for all positive integers $n$,

$\sum_{r=1}^{n} r(2r^2 - 3r - 1) = \frac{1}{2}n(n + 1)^2(n - 2)$

(b) Hence show that, for all positive integers $n$,

$\sum_{r=n}^{2n} r(2r^2 - 3r - 1) = \frac{1}{2}n(n - 1)(an + b)(cn + d)$

where $a$, $b$, $c$ and $d$ are integers to be determined.