Question

Find the sum of $n$ terms of the series whose $nth$ term is :
$n(n+3)$

Answer 1

$\sum n(n+3)$

$\sum n^2+3n$

$\sum n^2+3\sum n$

$\frac{n\left(n+1\right)(2n+1)}{6}+\frac{3n(n+1)}{2}$

$\frac{n(n+1)}{2}[\frac{2n+1}{3}+3]$