Question

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

Answer 1

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

$S_n=\sum T_n=\frac{n(n+1)}{2}+n+S_1$

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

$A_1=1-\frac{1}{3}$

$A_2=\frac{1}{2}-\frac{1}{4}$

$A_3=\frac{1}{3}-\frac{1}{5}$

$A_4=\frac{1}{4}-\frac{1}{6}$

$\dots \dots \dots \dots \dots \dots$

$\dots \dots \dots \dots \dots \dots$

$A_{n-1}=\frac{1}{n-1}-\frac{1}{n+1}$

$A_n=\frac{1}{n}-\frac{1}{n+2}$

$S_1=A_1+A_2+A_3+A_4+\dots +A_{n-1}+A_n=1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2}=\frac{3}{2}-\frac{2n+3}{(n+1)(n+2)}$

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