If the sum of the first ten terms of the series $\frac{1}{5}+\frac{2}{65}+\frac{3}{325}+\frac{4}{102

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

Professor PC Thomas Classes · Answer

(276) : $\frac{1}{5}+\frac{2}{65}+\frac{3}{325}+\frac{4}{1025}+\ldots$\ $\Rightarrow \quad t_{r}=\frac{r}{4 r^{4}+1}$\ $\Rightarrow \quad t_{r}=\frac{r}{\left(2 r^{2}+1ight)^{2}-(2 r)^{2}}$\ $\Rightarrow \quad t_{r}=\frac{1}{4} 	imes \frac{4 r}{\left(2 r^{2}+1-2 right)\left(2 r^{2}+1+2 right)}$\ $=\frac{1}{4}\left[\frac{1}{2 r^{2}+1-2 r}-\frac{1}{2 r^{2}+1+2 r}ight]$\ $\Rightarrow \sum_{r=1}^{10} t_{r}=\frac{1}{4} \sum_{r=1}^{10}\left(\frac{1}{2 r^{2}+1-2 r}-\frac{1}{2 r^{2}+1+2 r}ight)$ $$ =\frac{1}{4}\left[\begin{array}{l} \left(\frac{1}{1}-\frac{1}{5}ight)+\left(\frac{1}{5}-\frac{1}{13}ight)+\left(\frac{1}{13}-\frac{1}{25}ight) \ +\ldots\left(\frac{1}{181}-\frac{1}{221}ight) \end{array}ight] $$ $=\frac{1}{4} 	imes \frac{220}{221}=\frac{55}{221}=\frac{m}{n}$\ $	herefore \quad m+n=55+221=276$\