Question

If the sum of the first ten terms of the series $\frac{1}{5}+\frac{2}{65}+\frac{3}{325}+\frac{4}{1025}+\frac{5}{2501}+\dots$ is $\frac{m}{n}$, where $m$ and $n$ are co-prime numbers, then $m+n$ is equal to_____.

Answer 1

Answer: 276

$\frac{1}{5}+\frac{2}{65}+\frac{3}{325}+\frac{4}{1025}+\frac{5}{2501}+\dots \dots .$
$T_n=\frac{n}{4n^4+1}$
$=\frac{n}{{\left(2n^2+1\right)}^2-(2n{)}^2}=\frac{n}{\left(2n^2+2n+1\right)\left(2n^2-2n+1\right)}$
$=\frac{1}{4}\left[\frac{1}{2n^2-2n+1}-\frac{1}{2n^2+2n+1}\right]$
$S_{10}=\sum _{n=1}^{10}{T}_n=\frac{1}{4}\left[\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{13}+\dots \dots \cdot \frac{1}{200+20+1}\right]$
$=\frac{1}{4}\left[1-\frac{1}{221}\right]=\frac{1}{4}\times \frac{220}{221}-\frac{55}{221}=\frac{m}{n}$
$m+n=55+221=276$