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Evaluate the sum 6/(3^2 - 1) + 6/(4^2 - 1) + 6/(5^2 - 1) + 6/(6^2 - 1) + 6/(7^2 - 1) + ...

 Jun 7, 2022
 #1
avatar+124676 
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Note that we can write    1/ [ n^2 -1]   as   1/[ 2(n -1)] - 1/[2(n + 1)]

 

So we have

 

6  (  1 / [ 2(3-1)] - 1/[2(3 + 1)]    +  1/[2 (4-1)] - 1 /[2(4 + 1)]  + 1/[2(5-1)] - 1/[2(5 + 1) ]   +  ..........  )   =

 

6  [ 1/4 - 1/8 + 1/6 -  1/10 + 1/8 - 1/12 + 1/10 - 1/14 + 1/12 - 1/16 + 1/14 - 1/18 + ......] 

 

All the terms in red will "cancel"  and we are  left with

 

6  [ 1/4 + 1/6]  =

 

6 [ 10 / 24 ]   =

 

6[ 5/12]  =

 

5 / 2

 

cool cool cool

 Jun 7, 2022

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