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Compute \([\sum_{n=1}^{1000} \frac{1}{n^2 + n}]\)

 May 12, 2020
edited by Guest  May 12, 2020
 #1
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sum_(n=1)^1000 1/(n^2 + n) = 1000/1001

 May 12, 2020
 #2
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Note  that we can write  1 / [ n^2 + n]   as      1/ n  - 1 / [ n + 1 ]

 

So  we  can write this series  in this manner  :

 

[ 1/1 - 1/ 2]   +  [ 1/2 - 1/3]  + [ 1/3 - 1/4] + [ 1/4 - 1/5]  +  ...... + [ 1/998 - 1/999] + [ 1/999 - 1/1000] + (1/1000 - 1/1001)  =

 

1  + (1/2 - 1/2)  + (1/3 - 1/3) + (1/4 - 1/4) + .......+ (1/998 - 1/998) + (1/999 - 1/999) - (1/1000 -1/1000) - 1/1001 =

 

1  - 1/1001  =

 

[ 1001 - 1 ]  / 1001  =

 

1000 / 1001

 

 

cool cool cool

 May 12, 2020

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