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For every positive integer n, the nth term of sequence is gi

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For every positive integer n, the nth term of sequence is gi [#permalink] New post 30 Jun 2020, 09:17
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Question Stats:

60% (01:05) correct 40% (04:43) wrong based on 5 sessions
For every positive integer n, the nth term of sequence is given by an= 1/n - 1/(n+1). What is the sum of the first 100 terms?

(a) 1
(b) 0
(c) 25
(d) 99/100
(e) 100/101
[Reveal] Spoiler: OA

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Re: For every positive integer n, the nth term of sequence is gi [#permalink] New post 30 Jun 2020, 09:17
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Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
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Re: For every positive integer n, the nth term of sequence is gi [#permalink] New post 30 Jun 2020, 09:43
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Carcass wrote:
For every positive integer n, the nth term of sequence is given by an= 1/n - 1/(n+1). What is the sum of the first 100 terms?

(a) 1
(b) 0
(c) 25
(d) 99/100
(e) 100/101


term1 = 1/1 - 1/2
term2 = 1/2 - 1/3
term3 = 1/3 - 1/4
term4 = 1/4 - 1/5
.
.
.
term99 = 1/99 - 1/100
term100 = 1/100 - 1/101

So, the sum looks like this:

Sum = (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5) + .... + (1/99 - 1/100) + (1/100 - 1/101)

NOTICE THAT MOST TERMS CANCEL OUT

Sum = (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5) + .... + (1/99 - 1/100) + (1/100 - 1/101)
= 1/1 - 1/101
= 101/101 - 1/101
= 100/101
= E

Cheers,
Brent
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Re: For every positive integer n, the nth term of sequence is gi   [#permalink] 30 Jun 2020, 09:43
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