If the sum of the first ten terms of the series 1 5 + 2 65 + 3 325 + 4 1025 + 5 2501 + … . is m n ,…

For the series $\frac{1}{5}+\frac{2}{65}+\frac{3}{325}+\frac{4}{1025}+\frac{5}{2501}+\dots \dots$

$T_n=\frac{n}{4n^4+1}$

$=\frac{n}{{\left(2n^2+1\right)}^2-(2n{)}^2}=\frac{n}{\left(2n^2+2n+1\right)\left(2n^2-2n+1\right)}$

$=\frac{1}{4}\left[\frac{1}{2n^2-2n+1}-\frac{1}{2n^2+2n+1}\right]$

Now $S_{10}=\sum _{n=1}^{10}{ T}_n=\sum _{n=1}^{10}\frac{1}{4}\left[\frac{1}{2n^2-2n+1}-\frac{1}{2n^2+2n+1}\right]$

$=\frac{1}{4}\left[\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{13}+\frac{1}{13}\dots \dots \dots +\frac{1}{181}-\frac{1}{221}\right]$

$=\frac{1}{4}\left[1-\frac{1}{221}\right]=\frac{55}{221}=\frac{m}{n}$

Hence, $m+n=55+221=276$