Question

Prove that the sum of n terms of series $\mathrm{1.3.5}+\mathrm{3.5.7}+\mathrm{5.7.9}+....$ is equal to
$n(2n^3+8n^2+7n-2).$

Answer 1

$a_r=(2r-1)(2r+1)(2r+3)$

$=8r^3+12r^2-2r-3$

$S_n=\sum _{r=1}^n{a}_r$

$=8\sum _{r=1}^n{r}^3+12\sum _{r=1}^n{r}^2-\sum _{r=1}^{n}r-3n$

$=8{\left[\frac{n(n+1)}{2}\right]}^2+12\left[\frac{n(n+1)(2n+1)}{6}\right]-2\left[\frac{n(n+1)}{2}\right]-3$

$=n(2n^3+8n^2+7n-2)$.