Find the sum of each of the following series .
(1) $\displaystyle \sum^{2n}_{r=n}r(2r+3)$
(2) $\displaystyle \sum^{3n}_{r=1}r(r^2+1)$
$\displaystyle \sum^n_{r=1} {r^2} = \frac{n(n+1)(2n+1)}{6} $
$\displaystyle
\sum^n_{r=1} {r^3} = \frac{n^2(n+1)^2}{4}
$
$\displaystyle
\sum^n_{r=1} {r} =\frac{n(n+1)}{2}
$
Use it in your case