Question

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

Answer 1

The correct answer is 276
$T_r = \frac{r}{(2r^2)^2+1}$
$= \frac{r}{(2r^2+1)^2-(2r)^2}$
$= \frac{1}{4} \frac{4r}{(2r^2+2r+1)(2r^2-2r+1)}$
$S_{10} = \frac{1}{4} \sum\limits_{r=1}^{10} \left( \frac{1}{(2r^2 - 2r + 1)} - \frac{1}{(2r^2 + 2r + 1)} \right)$
$= \frac{1}{4} \left[ 1 - \frac{1}{5} + \frac{1}{5} - \frac{1}{13} + \ldots + \frac{1}{181} - \frac{1}{221} \right]$
⇒ $S_{10} = \frac{1}{4}. \frac{220}{221}$
$= \frac{55}{221} = \frac{m}{n}$
Hence, m+n = 276