Awesome q2a theme
0 votes
29 views
Find the following sum

$$\frac{1}{2^2 – 1} + \frac{1}{4^2 – 1} + \frac{1}{6^2 – 1} + … + \frac{1}{40^2 – 1}$$

(A) $\frac{20}{41}$

(B) $\frac{10}{41}$

(C) $\frac{10}{21}$

(D) $\frac{20}{21}$

(E) $1$
in Calculus by (3.6k points)
retagged by | 29 views

1 Answer

+1 vote
Best answer
$\frac{1}{a^2 - 1} = \frac{1}{(a+1)(a-1)} = \frac{1}{2} \left (\frac{1}{a-1} - \frac{1}{a+1} \right )$

So, I can write the given sum as :

$\frac{1}{2} \left [ \left (\frac{1}{2-1} - \frac{1}{2+1}  \right ) + \left (\frac{1}{4-1} - \frac{1}{4+1}  \right ) + \left (\frac{1}{6-1} - \frac{1}{6+1}  \right ) +……. + \left (\frac{1}{40-1} - \frac{1}{40+1}  \right ) \right ]$

$\frac{1}{2} \left [ \left (\frac{1}{1} - \frac{1}{3}  \right ) + \left (\frac{1}{3} - \frac{1}{5}  \right ) + \left (\frac{1}{5} - \frac{1}{7}  \right ) + \frac{1}{7} – ……. + \left (\frac{1}{39} - \frac{1}{41}  \right ) \right ]$

$\frac{1}{2} \left ( \frac{1}{1} – \frac{1}{41} \right)$ $(\because \frac{1}{a+1} = \frac{1}{(a+2)-1})$

$\frac{20}{41} $
by (357 points)
selected by
Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
Welcome to GATE CSE Doubts, where you can ask questions and receive answers from other members of the community.
9,200 questions
3,182 answers
14,686 comments
96,168 users