Question

Find the sum of $n$ terms of the series whose $n^{th}$ term is $\frac{n^4+2n^3+n^2-1}{n^2+n}$.

Answer 1

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

$S_n=\sum T_n=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}+(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\dots +\frac{1}{n}-\frac{1}{n+1})$

$S_n=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}+1-\frac{1}{n+1}=\frac{n(n+1)(n+2)}{3}-\frac{n}{n+1}$