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In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X [#permalink]

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31 Mar 2010, 11:47

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58% (02:19) correct
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In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X^(n+2) + X^(n+3) where x is a positive integer constant. For what value of n is the ratio of An to x(1+x(1+x(1+x(1+x)))) equal to X^5?

18. In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X^(n+2) + X^(n+3) where x is a positive integer constant. For what value of n is the ratio of An to x(1+x(1+x(1+x(1+x)))) equal to X^5?

(A) 8

(B) 7

(C) 6

(D) 5

(E) 4

note: An= A sub n

Can you explain this in detail? I tried expanding out the bottom equation and solving for X to equal x^5. Didnt really work out...

Its not so hard when you realize how can you solve it, but until that, you spent half of your life.

Oh no you don't. Work smart!

\(An = x^{n-1} + x^n + x^{n+1} + x^{n+2} + x^{n+3}\) e.g. \(A2 = x + x^2 + x^3 + x^4 + x^5\) Notice you can only take x common out of all these terms i.e. the smallest term \(x^{n - 1}\)

If \(\frac{An}{{x(1+x(1+x(1+x(1+x))))}} = x^5\), it means the part: (1+x(1+x(1+x(1+x)))) will get canceled from the num and den. Ignore it. From An, you will be able to take out \(x^6\) common so that \(\frac{x^6}{x}\) gives you \(x^5\) So smallest term must be \(x^6\) i.e. \(x^{n-1}\). Therefore, n = 7.
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"it means the part: (1+x(1+x(1+x(1+x)))) will get canceled from the num and den."

Regards, Subhash

\(A2 = x + x^2 + x^3 + x^4 + x^5\)

\(\frac{An}{{x(1+x(1+x(1+x(1+x))))}} = x^5\) Since the right side of the equation is just x^5, it means the entire expression: (1+x(1+x(1+x(1+x)))) should get canceled out which means we will get the same expression in the numerator as well. You don't need to do it. It is logical since otherwise, you will not get the reduced expression x^5. Also, you can see that you will get something like this in the numerator since the powers are increasing.

If you want to see it: \(A2= x( 1 + x + x^2 + x^3 + x^4) = x( 1 + x(1 + x + x^2 + x^3)) = x( 1 + x(1 + x( 1 + x + x^2))) = x( 1 + x(1 + x( 1 + x( 1 + x))))\)

Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X [#permalink]

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24 Jul 2012, 19:27

omg.. by the time u read and digest the question its 1 minut e
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I've failed over and over and over again in my life and that is why I succeed--Michael Jordan Kudos drives a person to better himself every single time. So Pls give it generously Wont give up till i hit a 700+

omg.. by the time u read and digest the question its 1 minut e

It certainly takes you a minute or even more to get through the question and digest it but after that, it takes you less than a minute to solve it. This is true for most GMAT questions. If you understand the question well, it takes you very little time to actually solve it. If you don't understand the question well, you could end up spending 20 mins on it.
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Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X [#permalink]

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31 Jul 2012, 01:49

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mohankumarbd wrote:

Is this a typical 700 level GMAT question? or just an off topic question? experts pls advice.

Who can tell you? If you ask all those who took the test if they ever saw such a question on a real test, you might get the real picture...

IMO, the chance is slim that such a question will appear on a real test. It is too technical, too lengthy to be done with plugging in numbers... Until now, I didn't get the feeling that GMAT wants to test just algebraic abilities. Not that this question needs some really advanced techniques, but it's above basics...
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PhD in Applied Mathematics Love GMAT Quant questions and running.

Is this a typical 700 level GMAT question? or just an off topic question? experts pls advice.

It is an algebra question that looks tricky but can be easily reasoned out. It will take you some time to understand the question but once you do, you can solve it quickly - pretty much like high level GMAT questions.
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Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X [#permalink]

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05 Sep 2012, 09:02

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I did it like it : as you can see x(1+x(1+x(1+x(1+x)))) X comes 5 times, therefore the max term will be X^5, in the question you see that you want to arrive at X^5 so it means that in the sum of X^n-1...x^n+3 the max term must be X^10 so that it can be x^5(x^5) therefore 10 = 3+n, n=7, timer indicate me 1min 53.

But definitely i had the answer, but i was unable to demonstrate it in that time, it would take more like 5 to 10 minutes.

Could someone please explain to me how that's an infinite sequence? That's what really threw me off.

The information that it is an infinite sequence doesn't have much to do with the question. You are given this only to tell you that n can take any positive integer value.

An = X^(n-1) + X^n + X^(n+1) + X^(n+2) + X^(n+3) tells you that the nth term is given by plugging in the value of n in this expression. A is not a sequence of 2 or 4 terms but infinite so n can take any value. We found out that the required relation holds when n is 7. We could have just as well got n = 10298 and that would have been fine too since A has infinite terms so any value for n is alright.
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