If the sum of the first ten terms of the series
$${1 \over 5} + {2 \over {65}} + {3 \over {325}} + {4 \over {1025}} + {5 \over {2501}} + \,\,....$$
is ${m \over n}$, where m and n are co-prime numbers, then m + n is equal to ______________.
Answer (integer)
276
Solution
<p>${T_r} = {r \over {{{(2{r^2})}^2} + 1}}$</p>
<p>$= {r \over {{{(2{r^2} + 1)}^2} - {{(2r)}^2}}}$</p>
<p>$= {1 \over 4}{{4r} \over {(2{r^2} + 2r + 1)(2{r^2} - 2r + 1)}}$</p>
<p>$${S_{10}} = {1 \over 4}\sum\limits_{r = 1}^{10} {\left( {{1 \over {(2{r^2} - 2r + 1)}} - {1 \over {(2{r^2} + 2r + 1)}}} \right)} $$</p>
<p>$$ = {1 \over 4}\left[ {1 - {1 \over 5} + {1 \over 5} - {1 \over {13}} + \,\,....\,\, + \,\,{1 \over {181}} - {1 \over {221}}} \right]$$</p>
<p>$$ \Rightarrow {S_{10}} = {1 \over 4}\,.\,{{220} \over {221}} = {{55} \over {221}} = {m \over n}$$</p>
<p>$\therefore$ $m + n = 276$</p>
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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