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Option $(A)$

$= \frac{1}{2^2-1} + \frac{1}{4^2-1} + ... + \frac{1}{40^2-1}$

$= \frac{1}{(2+1)(2-1)} + \frac{1}{(4+1)(4-1)} + ... +\frac{1}{(40+1)(40-1)}$

$= [\frac{1}{(2+1)(2-1)} + \frac{1}{(4+1)(4-1)} + ... +\frac{1}{(40+1)(40-1)}]\frac{2}{2}$

$= [\frac{2}{(2+1)(2-1)} + \frac{2}{(4+1)(4-1)} + ... +\frac{2}{(40+1)(40-1)}]\frac{1}{2}$

$= [\frac{(2+1)-(2-1)}{(2+1)(2-1)} + \frac{(4+1)-(4-1)}{(4+1)(4-1)} + ... +\frac{(40+1)-(40-1)}{(40+1)(40-1)}]\frac{1}{2}$

$= [\frac{1}{(2-1)} - \frac{1}{(2+1)} + \frac{1}{(4-1)} - \frac{1}{(4+1)} + ... +\frac{1}{(40-1)} - \frac{1}{(40+1)}]\frac{1}{2}$

$= [\frac{1}{1} - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + ... +\frac{1}{39} - \frac{1}{41}]\frac{1}{2}$

$= [1 - \frac{1}{41}]\frac{1}{2}$

$= \frac{20}{41}$
Answer:

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