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The value of the following expression :

$\left [ \frac{1}{2^{2}-1} \right ]+\left [ \frac{1}{4^{2}-1} \right ]+\left [ \frac{1}{6^{2}-1} \right ]+\dots\dots+\left [ \frac{1}{20^{2}-1} \right ]$ is :

  1. $\frac{10}{21}$
  2. $\frac{13}{27}$
  3. $\frac{15}{22}$
  4. $\frac{22}{15}$
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$\cfrac{1}{2^{2}-1}+\cfrac{1}{4^{2}-1}+\cfrac{1}{6^{2}-1}+\cdots+\cfrac{1}{20^{2}-1}$

$= \cfrac{1}{(2+1)(2-1)} + \cfrac{1}{(4+1)(4-1)} +\cfrac{1}{(6+1)(6-1)} +\cdots +\cfrac{1}{(20+1)(20-1)}$

$= \cfrac{2}{2}\left [\cfrac{1}{(2+1)(2-1)} + \cfrac{1}{(4+1)(4-1)}+\cfrac{1}{(6+1)(6-1)} +\cdots +\cfrac{1}{(20+1)(20-1)}\right]$

$= \cfrac{1}{2}\left [\cfrac{2}{(2+1)(2-1)} + \cfrac{2}{(4+1)(4-1)} +\cfrac{2}{(6+1)(6-1)} +\cdots +\cfrac{2}{(20+1)(20-1)}\right]$ $\require{cancel}$

$= \cfrac{1}{2}\left [\cfrac{(2+1)-(2-1)}{(2+1)(2-1)} + \cfrac{(4+1)-(4-1)}{(4+1)(4-1)} + \cfrac{(6+1)-(6-1)}{(6+1)(6-1)}+\cdots +\cfrac{(20+1)-(20-1)}{(20+1)(20-1)}\right]$

$=\cfrac{1}{2}\left [\cfrac{\cancel{(2+1)}}{\cancel{(2+1)}(2-1)}-\cfrac{\cancel{(2-1)}}{(2+1)\cancel{(2-1)}}+\cfrac{\cancel{(4+1)}}{\cancel{(4+1)}(4-1)}-\cfrac{\cancel{(4-1)}}{(4+1)\cancel{(4-1)}}+\cfrac{\cancel{(6+1)}}{\cancel{(6+1)}(6-1)}-\cfrac{\cancel{(6-1)}}{(6+1)\cancel{(6-1)}}+\cdots+\cfrac{\cancel{(20+1)}}{\cancel{(20+1)}(20-1)}-\cfrac{\cancel{(20-1)}}{(20+1)\cancel{(20-1)}}\right]$

$= \cfrac{1}{2}\left [\cfrac{1}{(2-1)} - \cfrac{1}{(2+1)} + \cfrac{1}{(4-1)} - \cfrac{1}{(4+1)} + \cfrac{1}{(6-1)} - \cfrac{1}{(6+1)}+\cfrac{1}{(8-1)} - \cfrac{1}{(8+1)}+\cdots +\cfrac{1}{(20-1)} - \cfrac{1}{(20+1)}\right]$

$= \cfrac{1}{2}\left [\cfrac{1}{1} - \cfrac{1}{3} + \cfrac{1}{3} - \cfrac{1}{5} + \cfrac{1}{5} - \cfrac{1}{7}+\cfrac{1}{7} - \cfrac{1}{9}+\cdots +\cfrac{1}{19} - \cfrac{1}{21}\right]$

$= \cfrac{1}{2}\left[\cfrac{1}{1} - \cfrac{1}{21}\right]$

$= \cfrac{1}{2}\left[ \cfrac{20}{21}\right]$

$= \cfrac{10}{21}$

Similar question: https://gateoverflow.in/358957/tifr2021-a-11
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