Question

If $S$ is sum of $\frac{5}{1^2{.2}^2.3}+\frac{8}{2^2{.3}^2.4}+\frac{11}{3^2.4^2.5}+\frac{14}{4^2{.5}^2.6}+.....\infty$ terms, then $4S$ is equal to

Answer 1

$\begin{array}{c}Tr=\frac{3r+2}{\left(r^2\right){\left(r+1\right)}^2\left(r+2\right)}\\ =\frac{r^2+3r+2-r^2}{r^2\left(r+1^2\right)\left(r+2\right)}\\ =\frac{1}{r^2\left(r+1\right)}-\frac{1}{{\left(r+1\right)}^2\left(r+2\right)}\\ T_1=\frac{1}{1\cdot 2}-\frac{1}{2^2\cdot 3}\\ T_2=\frac{1}{2^2\cdot 3}-\frac{1}{3^2\cdot 4}\\ T_3=\frac{1}{3^2\cdot 4}-\frac{1}{4^2\cdot 5}\\ T_n=\frac{1}{n^2\left(n+1\right)}-\frac{1}{{\left(n+1\right)}^2\left(n+2\right)}\\ S_n=\frac{1}{1\cdot 2}-\frac{1}{{\left(n+1\right)}^2\left(n+2\right)}\\ S_{\infty }=\frac{1}{2}-\frac{1}{\infty }\\ n\to \infty \\ S_{\infty }=1/2\\ Now,\\ 4S=\frac{4\times 1}{2}=2\end{array}$