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

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

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New post 19 Sep 2013, 13:34
Could someone please explain to me how that's an infinite sequence? That's what really threw me off.
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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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New post 19 Sep 2013, 21:48
Expert's post
mfabros wrote:
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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Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X [#permalink]

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New post 13 Sep 2014, 09:18
FYI guys, this is a Manhattan Review Turbocharge question.
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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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New post 10 Oct 2014, 03:26
It is actually very easy
Expand Brackets
we need An/(x+...+x^5) = x^5
An = x^5 (x+...+x^5)
An = x^6+....+x^10
only n = 7 satisfies..
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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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New post 15 Jun 2015, 14:15
I have solved it in this way:

The question asks to find n where
An/x(1+x(1+x(1+x(1+x)))) = x^5 (1)

Factorizing An= x^(n-2)(x(1+x(1+x(1+x(1+x)))) (2)

Replacing (2) in (1) we get: X^(n-2)=x^5-->X^n=x^7-->n=7

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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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New post 05 Jun 2016, 19:34
A different approach...

The maximum power of "x" in the denominator is 5 because "x" appears 5 times. For the ratio to be x^5, the numerator should have x^10. The highest power of "x" in the numerator is 10; so by comparison, n+3=10 => n = 7.
Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X   [#permalink] 05 Jun 2016, 19:34

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

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