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

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If A(n)=1/(n(n+1)) for all positive integers n, what is the [#permalink] New post 01 Sep 2013, 11:46
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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.

Last edited by Bunuel on 01 Sep 2013, 11:48, edited 1 time in total.
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Re: If A(n)=1/(n(n+1)) for all positive integers n, what is the [#permalink] New post 01 Sep 2013, 21:05
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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 =
[Reveal] Spoiler:
100/101
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Re: If A(n)=1/(n(n+1)) for all positive integers n, what is the [#permalink] New post 01 Sep 2013, 21:13
@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.
I hope that answers your question.

/SW
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Re: If A(n)=1/(n(n+1)) for all positive integers n, what is the [#permalink] New post 09 Sep 2013, 12:30
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
Re: If A(n)=1/(n(n+1)) for all positive integers n, what is the   [#permalink] 09 Sep 2013, 12:30
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