Question

Let sum the following series to n terms and infinity $\frac{1}{\mathrm{1.4.7}}+\frac{1}{\mathrm{4.7.10}}+\frac{1}{\mathrm{7.10.13}}+......$ be $\frac{1}{k}-\frac{1}{m(3n+1)(3n+4)},\frac{1}{24}$.Find $k-m-l$ ?

Answer 1

$\frac{1}{\mathrm{1.4.7}}+\frac{1}{\mathrm{4.7.10}}+\frac{1}{\mathrm{7.10.13}}+......$

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

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

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

$S_n=\frac{1}{6}[\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+\frac{1}{(3n+1)(3n+4)}]$

$=\frac{1}{6}[\frac{1}{4}-\frac{1}{(3n+1)(3n+4)}]$

Now $S_{\infty }=\frac{1}{24}$

$k=24,m=6,l=4$

$k-m-l=14$