Find the sum to n terms of the series whose nth term is given by n (n + 1) (n + 4).
Question:

Find the sum to $n$ terms of the series whose $n^{\text {th }}$ term is given by $n(n+1)(n+4)$.

Solution:

$a_{n}=n(n+1)(n+4)=n\left(n^{2}+5 n+4\right)=n^{3}+5 n^{2}+4 n$

$\therefore S_{n}=\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n} k^{3}+5 \sum_{k=1}^{n} k^{2}+4 \sum_{k=1}^{n} k$

$=\frac{n^{2}(n+1)^{2}}{4}+\frac{5 n(n+1)(2 n+1)}{6}+\frac{4 n(n+1)}{2}$

$=\frac{n(n+1)}{2} \mid \frac{n(n+1)}{2}+\frac{5(2 n+1)}{3}+4$

$=\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+\frac{5(2 n+1)}{3}+4\right]$

$=\frac{n(n+1)}{2}\left[\frac{3 n^{2}+23 n+34}{6}\right]$

$=\frac{n(n+1)\left(3 n^{2}+23 n+34\right)}{12}$