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For every positive integer n, the nth term of sequence is gi
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30 Jun 2020, 09:17
30 Jun 2020, 09:17
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78% (01:19) correct
21% (02:58) wrong based on 37 sessions
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For every positive integer n, the nth term of sequence is given by \(a_n= \frac{1}{n} - \frac{1}{n+1}\). What is the sum of the first 100 terms?
(A) 1
(B) 0
(C) 25
(D) 99/100
(E) 100/101
(A) 1
(B) 0
(C) 25
(D) 99/100
(E) 100/101
Re: For every positive integer n, the nth term of sequence is gi
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30 Jun 2020, 09:17
30 Jun 2020, 09:17
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Expert Reply
Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
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Re: For every positive integer n, the nth term of sequence is gi
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30 Jun 2020, 09:43
30 Jun 2020, 09:43
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Carcass wrote:
For every positive integer n, the nth term of sequence is given by \(a_n= \frac{1}{n} - \frac{1}{n+1}\). What is the sum of the first 100 terms?
(A) 1
(B) 0
(C) 25
(D) 99/100
(E) 100/101
(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
