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f(n) = 1/n 1/(n+1) what is the sum of the first 100 integers

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f(n) = 1/n 1/(n+1) what is the sum of the first 100 integers [#permalink]

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New post 21 May 2004, 06:48
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

f(n) = 1/n – 1/(n+1) what is the sum of the first 100 integers?
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New post 21 May 2004, 09:03
Just a guess! :lol:

1/n - 1/(n+1) = 1 / n(n+1)
This is a diminishing sequence. I added few decimals in the bgigin and found that as n increases very small incremental increase results in 1/n(n+1) from previous number.
There is a formula for this but I dont remember.

0.5 + 0.1666 + 0.0833 + 0.05 + ....
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Re: question [#permalink]

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New post 21 May 2004, 09:07
S(n) = f(1) + ...+ f(n)
f(1) = 1- (1/2)
f(2) = (1/2)-(1/3)
.
.
.
f(n) = (1/n)-(1/n+1)

S(100) = n/(n+1) = 1-(1/101) = 100/101
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thanks [#permalink]

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New post 21 May 2004, 16:07
thank hallelujah :lol: , u seem to know everything so i just copy your answers to every problem and review them later
thanks   [#permalink] 21 May 2004, 16:07
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f(n) = 1/n 1/(n+1) what is the sum of the first 100 integers

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