GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 22 Oct 2019, 02:25 GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  If A(n)=1/(n(n+1)) for all positive integers n, what is the

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

Hide Tags

Intern  Joined: 08 Mar 2013
Posts: 18
If A(n)=1/(n(n+1)) for all positive integers n, what is the  [#permalink]

Show Tags

1
8 00:00

Difficulty:

(N/A)

Question Stats: 100% (01:09) correct 0% (00:00) wrong based on 13 sessions

HideShow timer Statistics

So I'm working throu Manhattan GMAT advanced quant book and got stuck on this problem:

If A(n)=1/(n(n+1)) for all positive integers n, what is the sum of the first 100 elements of An?

The book calculates it by deriving a formula based on pattern by calculating several members of the series and examining changes in the numerator and denominator of the sum of each member + each previous member of the series. That derives to SUM=n/(n+1)

I tried it by using the average formula, AVE=SUM/N and I get a different result. Here're my calculations:

Ave=(A(1)+A(100))/2

A(1)=1/2
A(100)=1/(100(101))=1/10100
Ave=(1/2+1/10100)/2, which ends up being 5051/20200 (unless I made a mistake somewhere along the line).

From there on, based on the Ave formula:
N=100 (first 100 elements of the series)
SUM=AVE*N=5051*100/20200=5051/202

That, clearly doesn't match the answer derived by the method presented in the book!

So did I make mistake in calculations or am I just missing something here?

PS. I searched for this question on the board/google, and nothing came up, but still my apologies if I double-posted something that been discussed previously.

Originally posted by Dixon on 01 Sep 2013, 12:46.
Last edited by Bunuel on 01 Sep 2013, 12:48, edited 1 time in total.
RENAMED THE TOPIC.
Intern  Joined: 31 Jan 2013
Posts: 17
Schools: ISB '15
WE: Consulting (Energy and Utilities)
Re: If A(n)=1/(n(n+1)) for all positive integers n, what is the  [#permalink]

Show Tags

6
1
The way I approached:

A(n)=1/(n*n-1)= (1/n)-(1/n+1)

A(1) = 1 - 1/2
A(2) = 1/2 - 1/3
A(3) = 1/3 - 1/4
'
'

A(100) = 1/100 - 1/101

Sum of first 100 terms = 1-1/2+1/2-1/3+1/3-1/4 ......+1/100-1/101

= 1-1/101 =
100/101
General Discussion
Intern  Joined: 31 Jan 2013
Posts: 17
Schools: ISB '15
WE: Consulting (Energy and Utilities)
Re: If A(n)=1/(n(n+1)) for all positive integers n, what is the  [#permalink]

Show Tags

@Dixon, As I understand, you can apply the formula Sum=A1+An/2*n only when the series is evenly distributedm, which is not our case.

/SW
Retired Moderator P
Joined: 22 Aug 2013
Posts: 1428
Location: India
Re: If A(n)=1/(n(n+1)) for all positive integers n, what is the  [#permalink]

Show Tags

1
Dixon wrote:
So I'm working throu Manhattan GMAT advanced quant book and got stuck on this problem:

If A(n)=1/(n(n+1)) for all positive integers n, what is the sum of the first 100 elements of An?

The book calculates it by deriving a formula based on pattern by calculating several members of the series and examining changes in the numerator and denominator of the sum of each member + each previous member of the series. That derives to SUM=n/(n+1)

I tried it by using the average formula, AVE=SUM/N and I get a different result. Here're my calculations:

Ave=(A(1)+A(100))/2

A(1)=1/2
A(100)=1/(100(101))=1/10100
Ave=(1/2+1/10100)/2, which ends up being 5051/20200 (unless I made a mistake somewhere along the line).

From there on, based on the Ave formula:
N=100 (first 100 elements of the series)
SUM=AVE*N=5051*100/20200=5051/202

That, clearly doesn't match the answer derived by the method presented in the book!

So did I make mistake in calculations or am I just missing something here?

PS. I searched for this question on the board/google, and nothing came up, but still my apologies if I double-posted something that been discussed previously.

Hi Dixon

the formula that you are applying is applicable to an AP (arithmetic progression).. AP is one where difference between any two consecutive values is same throughout the series.. consider the following series:

4, 7, 10, 13, 16, 19... and so on.. you can see that in this series each term is 3 more than the previous term.. such a progession is called an AP.. now observe how interesting this is! sum of first and last term is (4+19) = 23, sum of second and second last term = (7+16) = 23, sum of (10+13) is also 23...

so what you are doing in the question that you posted is that you are considering the sequence as an AP which it is not..... so this has to be done using the stated method in Manhattan only...
I hope I have made some sense
Non-Human User Joined: 09 Sep 2013
Posts: 13385
Re: If A(n)=1/(n(n+1)) for all positive integers n, what is the  [#permalink]

Show Tags

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________ Re: If A(n)=1/(n(n+1)) for all positive integers n, what is the   [#permalink] 07 Oct 2018, 17:47
Display posts from previous: Sort by

If A(n)=1/(n(n+1)) for all positive integers n, what is the

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne  