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Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X^(n+2) + X

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Intern  Joined: 27 Aug 2014
Posts: 1
Schools: Haas '17
Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X^(n+2) + X  [#permalink]

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FYI guys, this is a Manhattan Review Turbocharge question.
Manager  Joined: 14 Sep 2014
Posts: 86
WE: Engineering (Consulting)
Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X^(n+2) + X  [#permalink]

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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..
Intern  Joined: 27 Apr 2014
Posts: 2
Concentration: General Management, Leadership
Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X^(n+2) + X  [#permalink]

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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
Intern  Joined: 01 Jun 2016
Posts: 1
Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X^(n+2) + X  [#permalink]

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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.
Manager  B
Joined: 31 Dec 2016
Posts: 65
Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X^(n+2) + X  [#permalink]

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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 [ 723.33 KiB | Viewed 869 times ]

Senior Manager  P
Joined: 29 Jun 2017
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Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X^(n+2) + X  [#permalink]

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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
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Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X^(n+2) + X  [#permalink]

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KarishmaB wrote:
subhashghosh wrote:
Hi Karishma

What is the meaning of this ?

"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:
$$A2= x( 1 + x + x^2 + x^3 + x^4) = x( 1 + x(1 + x + x^2 + x^3)) = x( 1 + x(1 + x( 1 + x + x^2))) = x( 1 + x(1 + x( 1 + x( 1 + x))))$$

Similarly A7 $$= x^6( 1 + x(1 + x( 1 + x( 1 + x))))$$

So $$\frac{A7}{{x(1+x(1+x(1+x(1+x))))}} = \frac{x^6( 1 + x(1 + x( 1 + x( 1 + x))))}{x( 1 + x(1 + x( 1 + x( 1 + x))))}= x^5$$

KarishmaB Can you explain how to do this step: 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))))

I don't get it! (

Tks!!!
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Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X^(n+2) + X  [#permalink]

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Loved the question. Solve it in under 1.5 min. Tricky. If one can understand the pattern, anything more than a minute is waste.
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Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X^(n+2) + X  [#permalink]

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subhashghosh wrote:
Hi Karishma

What is the meaning of this ?

"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:
$$A2= x( 1 + x + x^2 + x^3 + x^4) = x( 1 + x(1 + x + x^2 + x^3)) = x( 1 + x(1 + x( 1 + x + x^2))) = x( 1 + x(1 + x( 1 + x( 1 + x))))$$

Similarly A7 $$= x^6( 1 + x(1 + x( 1 + x( 1 + x))))$$

So $$\frac{A7}{{x(1+x(1+x(1+x(1+x))))}} = \frac{x^6( 1 + x(1 + x( 1 + x( 1 + x))))}{x( 1 + x(1 + x( 1 + x( 1 + x))))}= x^5$$

Quote:
Hi! Can you please explain me how to do this step: 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))))

You are just taking out x's common from the remaining terms one after the other.

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))))
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Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X^(n+2) + X  [#permalink]

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_________________ Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X^(n+2) + X   [#permalink] 21 Aug 2019, 08:35

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