Question

Find the sum of the series whose $n^{th}$ term is :
$2n^3+3n^2-1$

Answer 1

$n^{th}$ term is $t_n=2n^3+3n^2-1$

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

$={\sum }_{k=1}^{n}2k^3+{\sum }_{k=1}^{n}3k^2-{\sum }_{k=1}^{n}1$

$=2{\sum }_{k=1}^n{k}^3+3{\sum }_{k=1}^n{k}^2-{\sum }_{k=1}^{n}1$

$=2\left(\frac{n^2(n+1{)}^2}{4}\right)+3\left(\frac{n(n+1)(2n+1)}{6}\right)-n$

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

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

$=\frac{n^4+n^2+2n^3+2n^3+3n^2+n-2n}{2}=\frac{n^4+4n^3+4n^2-n}{2}$