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For the sequence a1, a2, a3 ... an, an is defined by a_n=1/n−1/(n+1)
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20 Feb 2017, 02:01
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65% (01:53) correct 35% (02:29) wrong based on 130 sessions
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For the sequence a1, a2, a3 ... an, an is defined by \(a_n=\frac{1}{n}−\frac{1}{(n+1)}\) for each integer n≥1. What is the sum of the first 100 terms of this sequence? A. 100/99 B. 101/100 C. 100/101 D. 99/100 E. 999/1010
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For the sequence a1, a2, a3 ... an, an is defined by a_n=1/n−1/(n+1)
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20 Feb 2017, 02:20
Bunuel wrote: For the sequence a1, a2, a3 ... an, an is defined by \(a_n=\frac{1}{n}−\frac{1}{(n+1)}\) for each integer n≥1. What is the sum of the first 100 terms of this sequence?
A. 100/99 B. 101/100 C. 100/101 D. 99/100 E. 999/1010 \(a_n=\frac{1}{n}−\frac{1}{(n+1)}\) \(a_1 + a_2 + a_3 + a_4 + ......+ a_99 + a_100=\frac{1}{1}−\frac{1}{2}+\frac{1}{2}−\frac{1}{3}+\frac{1}{3}−\frac{1}{4}+\frac{1}{4}−\frac{1}{5}+.....+ \frac{1}{99}−\frac{1}{100}+\frac{1}{100}−\frac{1}{101}\) Everything in between 1/1 and (1/101) gets cancelled out. \(a_1 + a_2 + a_3 + a_4 + ......+ a_99 + a_100= \frac{1}{1}−\frac{1}{101} = \frac{100}{101}\) Hence Option C is correct. Hit Kudos if you liked it




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Re: For the sequence a1, a2, a3 ... an, an is defined by a_n=1/n−1/(n+1)
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20 Feb 2017, 02:27
C Coz a1= 1 1/2 a2 = 1/2 1/3 a100= 1/100  1/101 Adding all d above we get 11/101 = 100/101 Sent from my HM 1S using GMAT Club Forum mobile app



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Re: For the sequence a1, a2, a3 ... an, an is defined by a_n=1/n−1/(n+1)
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21 Jan 2018, 14:24
Just curious why the each term would not be included in parenthesis? Wouldn't this change the order of operations and the answer?
(A1)+(a2)... 1/2+((1/2)(1/3))+((1/3)(1/4))...



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For the sequence a1, a2, a3 ... an, an is defined by a_n=1/n−1/(n+1)
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30 Nov 2018, 07:43
I didn't pick up on the pattern while calculating a1, a2, a3; however, I did a couple of sums and picked up on the pattern then: a1=1/2 a1+a2=2/3 a1+a2+a3=3/4
so a1+a2+....+a100=100/100+1=100/101



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Re: For the sequence a1, a2, a3 ... an, an is defined by a_n=1/n−1/(n+1)
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26 Jun 2019, 08:51
BunuelAre you sure this is a Veritas Prep question? I am fairly certain that this question is actually from the GMAT Prep EP2 or a pretty blatant copy of an OG question. https://gmatclub.com/forum/foreverypo ... 17215.htmlBest regards, Chris
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Re: For the sequence a1, a2, a3 ... an, an is defined by a_n=1/n−1/(n+1)
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26 Jun 2019, 09:12
100/101 > a1= 1/1 1/2 a2 = 1/2 1/3 a3 = 1/3 1/4 a4 = 1/4 1/5 . . . a100=1/1001/101
the second term of the previous term and the first term of the present term get cancelled out (e.g 1/2 from a1 and 1/2 from a3 get cancelled out) ... leaving only 1/1 and 100/101
Thus, 1/11/101 = 101/1011/101 = 100/101



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Re: For the sequence a1, a2, a3 ... an, an is defined by a_n=1/n−1/(n+1)
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01 Jul 2019, 17:14
Bunuel wrote: For the sequence a1, a2, a3 ... an, an is defined by \(a_n=\frac{1}{n}−\frac{1}{(n+1)}\) for each integer n≥1. What is the sum of the first 100 terms of this sequence?
A. 100/99 B. 101/100 C. 100/101 D. 99/100 E. 999/1010 Let’s calculate the first few terms: a1 = 1/1  1/2 a2 = 1/2  1/3 a3 = 1/3  1/4 a4 = 1/4  1/5 Summing the first 4 terms of the sequence, we have 1  1/5 because all the middle numbers cancel. Thus, we see that only the first and last term are left. So the sum of the first 100 terms would be 1  1/101 = 100/101. Answer: C
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Re: For the sequence a1, a2, a3 ... an, an is defined by a_n=1/n−1/(n+1)
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01 Jul 2019, 17:14






