If the sum of the first ten terms of the series 15+265+3325+41025+52501+ is mn, where m,n …

Question

If the sum of the first ten terms of the series $\dfrac15+\dfrac{2}{65}+\dfrac{3}{325}+\dfrac{4}{1025}+\dfrac{5}{2501}+\cdots$ is $\dfrac{m}{n}$, where $m,n$ are coprime, then $m+n$ is equal to \_\_\_\_\_\_\_\_.

Answer: 276

Accepted Answer

The denominators are $4r^4+1$, so $T_r = \dfrac{r}{4r^4+1} = \dfrac14\left[\dfrac{1}{2r^2-2r+1}-\dfrac{1}{2r^2+2r+1}\right]$, a telescoping form.

$$S_{10} = \frac14\left(1-\frac{1}{221}\right) = \frac{55}{221} = \frac mn \implies m+n = 276.$$

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