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

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New post 09 Aug 2018, 07:20
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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New post 20 Aug 2018, 00:24
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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New post 20 Aug 2018, 08:11
VeritasKarishma 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 \(\frac{A7}{{x(1+x(1+x(1+x(1+x))))}} = \frac{x^6( 1 + x(1 + x( 1 + x( 1 + x))))}{x( 1 + x(1 + x( 1 + x( 1 + x))))}= x^5\)


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 &nbs [#permalink] 20 Aug 2018, 08:11

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