Question

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

Answer 1

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