Question

The sum of the first 10 terms of the series $\frac{7}{2^2 5^2}+\frac{13}{5^2 8^2}+\frac{19}{8^2 11^2}+\ldots \ldots \ldots$ is $\frac{ m }{ n }$ the find the value of $(n-12 m)$ ?

Answer 1

$S =\frac{7}{2^2 5^2}+\frac{13}{5^2 8^2}+\frac{19}{8^2 11^2}+\ldots \ldots \ldots$
$3 S =\frac{21}{2^2 5^2}+\frac{39}{5^2 8^2}+\frac{57}{8^2 11^2}+ $
$3 S =\displaystyle\sum_{ r =1}^{10} \frac{(3 r +2)^2-(3 r -1)^2}{(3 r -1)^2(3 r +2)^2}$
$3 S =\displaystyle\sum_{ r =1}^{10}\left(\frac{1}{(3 r -1)^2}-\frac{1}{(3 r +2)^2}\right. $
$3 S =\frac{1}{2^2}-\frac{1}{2^{10}} $
$3 S =\frac{2^8-1}{2^{10}} $
$S =\frac{85}{1024}=\frac{ m }{ n } .$
$\text { Hence, }( n -12 m )=4$