Question

Sum the following series to n terms $\mathrm{1.5.9}+\mathrm{2.6.10}+\mathrm{3.7.11}+...$.

Answer 1

General term$=n(n+4)(n+8)$

$=n^3+12n^2+32n$

Sum to $n$ terms $=\sum n^3+12\sum n^2+32\sum n$

$=\frac{n^2{(n+1)}^2}{4}+12\frac{\left(n\right)(n+1)(2n+1)}{6}+\frac{32n(n+1)}{2}$

$=\frac{n(n+1)}{4}[n^2+17n+72]$

$=\frac{n(n+1)(n+8)(n+9)}{4}$