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# For each integer n > 1, let A(n) denote the sum

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GRE Prep Club Legend
Joined: 07 Jun 2014
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GRE 1: Q167 V156
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For each integer n > 1, let A(n) denote the sum [#permalink]  12 May 2016, 04:33
Expert's post
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Question Stats:

82% (00:56) correct 17% (00:39) wrong based on 29 sessions
For each integer n > 1, let A(n) denote the sum of the integers from 1 to n. For example, A(100) = 1 + 2 + 3 + … + 100 = 5,050. What is the value of A(200)?

A. 10,100
B. 15,050
C. 15,150
D. 20,100
E. 21,500

Practice Questions
Question: 8
Page: 62
[Reveal] Spoiler: OA

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Sandy
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GRE Prep Club Legend
Joined: 07 Jun 2014
Posts: 4809
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 120

Kudos [?]: 1913 [0], given: 397

Re: For each integer n > 1, let A(n) denote the sum [#permalink]  12 May 2016, 04:40
Expert's post
Explanation

In the question, you are given that A(n) is equal to the sum of the integers from 1 to n, so A(200) = 1 + 2 + 3 +…+100 +101 +102 +103 +…+200. In order to be able to use the given value of A(100) = 5,050, you can rewrite the sum as:

$$A(200) = A(100)+ 101 + 102 + 103 ...............+ 200$$
$$= A(100)+ (100 +1) + (100 + 2) + (100 + 3)............... (100 + 100)$$
$$= A(100)+ (1+ 2+ ................ 100) + 100*100$$
$$= 5050 + 5050 + 10000$$
$$= 20100.$$

Thus the correct answer is Choice D.
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Sandy
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Re: For each integer n > 1, let A(n) denote the sum [#permalink]  23 Sep 2016, 17:05
In this question, ETS doesn't quite explain how they find it that well. The sum of positive integers equation is n(n+1)/2 If you'd like, you could solve this problem that way and still get the correct answer, D.

In the ETS description, they write this:
Rewrite the sum as:
A(200) = A(100) + 101 +102+103... +200
=A(100) + (100+1) + (100+2) ... +100+100 (that is showing dividing the final sum of 200 into two parts so that it can be made into two A(100)s.)
=A(100) + (1 +2+3... +100) (100)(100) (this was confusing for me to see, so I figure I'd explain it. Here, they've factored out 100 from the sum of 101 to 200, and thus factoring it out would multiply it by the remainder, in this case, 100.)
=A(100) + A(100) + (100)(100)
=20,100

Then again, one could just use the sum of positive integers formula for 1 to 200. ((200)(200+1))/2 which also comes to 20,100.

Hope this helps, fellow gre preppers!
Intern
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Re: For each integer n > 1, let A(n) denote the sum [#permalink]  26 Sep 2016, 03:44
How is "For each integer n > 1" helpful to the question? I spent far too much time trying to decide what that part of the question meant when the remainder of the information was sufficient to correctly answer the question. It would seem that the sum of each integer greater than 1 from 1 to 200 is actually 2 to 200...so the answer would actually be 20,099...
Intern
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Re: For each integer n > 1, let A(n) denote the sum [#permalink]  18 Dec 2016, 19:18
I don't really understand these explanations but here is how I thought about it. In the question they already tell you that 1+....100 is 5050. So A(200) must have two lots of 5050 as well as 100 lots of 100. So 5050+5050+10000 is 20100. You don't really need to do any difficult arithmetic if you think about it that way.
GRE Instructor
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Re: For each integer n > 1, let A(n) denote the sum [#permalink]  12 Jun 2019, 09:22
Expert's post
sandy wrote:
For each integer n > 1, let A(n) denote the sum of the integers from 1 to n. For example, A(100) = 1 + 2 + 3 + … + 100 = 5,050. What is the value of A(200)?

A. 10,100
B. 15,050
C. 15,150
D. 20,100
E. 21,500

Practice Questions
Question: 8
Page: 62

GIVEN: 1 + 2 + 3 + … + 100 = 5,050

A(200) = 1 + 2 + 3 + ....+ 99 + 100 + 101 + 102 + 103 + ......+ 199 + 200
= 1 + 2 + 3 + ....+ 99 + 100 + (100 + 1) + (100 + 2) + (100 + 3) + ....(100 + 99) + (100 + 100)
= (1 + 2 + 3 + ....+ 99 + 100) + (1 + 2 + 3 + ....+ 99 + 100) + 100(100)
= (5,050) + (5,050) + (10,000)
= 20,100

Cheers,
Brent
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Brent Hanneson – Creator of greenlighttestprep.com

Re: For each integer n > 1, let A(n) denote the sum   [#permalink] 12 Jun 2019, 09:22
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