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

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Intern
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Joined: 27 Aug 2014
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Schools: Haas '17
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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Manager
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Joined: 14 Sep 2014
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Kudos [?]: 35 [0], given: 51

WE: Engineering (Consulting)
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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Intern
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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

regards

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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.

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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 12 Aug 2017, 17:05
I think the question was harder because the original person didn't write it in clear notation. Using Bunnel's look makes it solvable.

I didn't solve here but I am explaining the math for those that don't get why when you remove an X^n-1 that it looks the way it does.
Attachments

how to do the math.png
how to do the math.png [ 723.33 KiB | Viewed 151 times ]

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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 Aug 2017, 12:20
solve the given x(1+x(1+x(1+x(1+x)))
count the min power of x , which will be x^1
now the An/ expression = x^5
means x^1 x x^ 5 = x ^ 6 is min power of x
now see nth term.. least power expression is x^(n-1)
n-1=6, n=7
Answer is B
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Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X   [#permalink] 13 Aug 2017, 12:20

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