Question

Sum the following series to n terms $\mathrm{1.4.7}+\mathrm{4.7.10}+\mathrm{7.10.13}+....$.

Answer 1

General term$=(3n-2)(3n+1)(3n+4)$

$=27n^3+27n^2-18n-8$

Sum to $n^{th}$ term $=27\sum n^3+27\sum n^2+18\sum n-8$

$=27\frac{n^2{(n+1)}^2}{4}+27\frac{\left(n\right)(2n+1)(n+1)}{6}-18\frac{\left(n\right)(n+1)}{2}-8$

$=n(n+1)[\frac{27(n^2+n)}{4}+\frac{27(2n+1)}{6}-17]$