Question

Find the sum of $n$ terms of the series whose $n^{th}$ term is
$n^2(2n+3)$.

Answer 1

$T_n=n^2(2n+3)=2n^3+3n^2$

Let $S_n$ denotes the sum of the $n$ terms of the given series, then

$S_n=\sum _{k=1}^n{T}_k=\sum _{k=1}^n(2k^3+3k^2)$

$=2\left(\sum _{k=1}^n{k}^3\right)+3\left(\sum _{k=1}^n{k}^2\right)$

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

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

$=\frac{n(n+1)(n^2+3n+1)}{2}$