Question

If sum of first $n$ terms of an A.P. is $S_n=3n^2-2n,$ then the value of $\sum _{n=1}^{\infty }\left(\frac{21}{(S_n{S}_{n+2}+S_{n-1}{S}_{n+1})-(S_n{S}_{n+1}+S_{n-1}{S}_{n+2})}\right)$ is equal to

Answer 1

$\sum _{n=1}^{\infty }\left(\frac{21}{(S_n{S}_{n+2}+S_{n-1}{S}_{n+1})-(S_n{S}_{n+1}+S_{n-1}{S}_{n+2})}\right)$
$=\sum _{n=1}^{\infty }\left(\frac{21}{S_n(S_{n+2}-S_{n+1})-S_{n-1}(S_{n+2}-S_{n+1})}\right)$
$=\sum _{n=1}^{\infty }\left(\frac{21}{(S_{n+2}-S_{n+1})(S_n-S_{n-1})}\right)$
$=\sum _{n=1}^{\infty }\left(\frac{21}{T_{n+2}\cdot T_n}\right)$

$\therefore T_n=S_n-S_{n-1}=3n^2-2n-(3{(n-1)}^2-2(n-1))=6n-5$
$T_{n+2}=6n+7$

Now,
$\sum _{n=1}^{\infty }\frac{21}{T_{n+2}\cdot T_n}=\sum _{n-1}^{\infty }\frac{21}{(6n+7)(6n-5)}=\frac{21}{12}\sum _{n=1}^{\infty }(\frac{1}{6n-5}-\frac{1}{6n+7})$
$=\frac{21}{12}[(\frac{1}{1}-\frac{1}{13})+(\frac{1}{7}-\frac{1}{19})+(\frac{1}{13}-\frac{1}{25})+(\frac{1}{19}-\frac{1}{31})+\dots ]$
$=\frac{21}{12}\times \frac{8}{7}=2$