Question

Sum the series to $n$ terms $\mathrm{2.4.6}+\mathrm{4.6.8}+\mathrm{6.8.10}+....$

Answer 1

$\mathrm{2.4.6}+\mathrm{4.6.8}+.....$

$a_n=2n\left(2n+2\right)\left(2n+4\right)$

$=8n\left(n+1\right)\left(n+2\right)$

$sumoftheseries=\sum _{n=1}^{n}an$

$=\sum 8n\left(n+1\right)\left(n+2\right)$

$=8\sum n\left(n^2+3n+2\right)$

$=8\left[\sum n^3+3\sum n^2+2\sum n\right]$

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

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

$=2n\left(n+1\right)\left[n^2+5n^2+6\right]$

$=2n\left(n+1\right)\left(n+2\right)\left(n+3\right)$