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Compute

SOLVED!

\(\sum_{n=1}^{1000} \frac{1}{n^2 + n}.\)

 Jun 7, 2019
edited by FlyEaglesFly  Jun 7, 2019
 #1
avatar+6026 
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\(\dfrac{1}{n^2+n}=\dfrac{1}{n(n+1)} = \dfrac{A}{n}+\dfrac{B}{n+1}\\~\\ A+B=0,~A=1,~B=-1\\~\\ \dfrac{1}{n^2+n} = \dfrac{1}{n}-\dfrac{1}{n+1}\)

 

\(\sum \limits_{n=1}^{1000}~\left(\dfrac 1 n - \dfrac{1}{n+1}\right) = \\~\\ \left(1 - \dfrac 1 2\right) + \left(\dfrac 1 2-\dfrac 1 3\right) + \left(\dfrac 1 3 - \dfrac 1 4\right) + \dots + \left(\dfrac{1}{1000}-\dfrac{1}{1001}\right) = \\~\\ 1-\dfrac{1}{1001} = \dfrac{1000}{1001}\)

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 Jun 7, 2019
 #2
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Thanks

 Jun 8, 2019

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