Question

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

Answer 1

Answer: 0004

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