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

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In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X [#permalink]  31 Mar 2010, 10:47
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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?

A. 8
B. 7
C. 6
D. 5
E. 4
[Reveal] Spoiler: OA
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Re: Tough Problem Solving Infinite Sequence Problem [#permalink]  31 Mar 2010, 11:33
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ksharma12 wrote:
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...

$$x^5=\frac{x^{(n-1)}+x^n+x^{(n+1)}+x^{(n+2)}+x^{(n+3)}}{x(1+x(1+x(1+x(1+x))))}$$

--> $$x^6(1+x(1+x(1+x(1+x))))=x^{(n-1)}+x^n+x^{(n+1)}+x^{(n+2)}+x^{(n+3)}$$

--> take $$x^{(n-1)}$$ out of the brackets

--> $$x^6(1+x(1+x(1+x(1+x))))=x^{(n-1)}(1+x+x^2+x^3+x^4)$$

--> $$x^6(1+x(1+x(1+x(1+x))))=x^{(n-1)}(1+x(1+x+x^2+x^3))$$

--> $$x^6(1+x(1+x(1+x(1+x))))=x^{(n-1)}(1+x(1+x(1+x+x^2)))$$

--> $$x^6(1+x(1+x(1+x(1+x))))=x^{(n-1)}(1+x(1+x(1+x(1+x))))$$

--> $$x^6=x^{(n-1)}$$ --> $$n-1=6$$ --> $$n=7$$

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Re: Tough Problem Solving Infinite Sequence Problem [#permalink]  31 Mar 2010, 12:19
great explanation. Thanks. +1 to you
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Re: Tough Problem Solving Infinite Sequence Problem [#permalink]  01 Apr 2010, 10:41
Is it a GMAT level question??????
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Re: Tough Problem Solving Infinite Sequence Problem [#permalink]  02 Feb 2011, 07:32
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Its not so hard when you realize how can you solve it, but until that, you spent half of your life.
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Re: Tough Problem Solving Infinite Sequence Problem [#permalink]  02 Feb 2011, 20:37
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craky wrote:
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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Re: Tough Problem Solving Infinite Sequence Problem [#permalink]  04 Feb 2011, 05:00
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
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Re: Tough Problem Solving Infinite Sequence Problem [#permalink]  04 Feb 2011, 07:20
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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 $$\frac{A7}{{x(1+x(1+x(1+x(1+x))))}} = \frac{x^6( 1 + x(1 + x( 1 + x( 1 + x))))}{x( 1 + x(1 + x( 1 + x( 1 + x))))}= x^5$$
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Re: Tough Problem Solving Infinite Sequence Problem [#permalink]  04 Feb 2011, 07:26
$$A_n=x^{(n-1)}(1+x+x^2+x^3+x^4)$$
$$x(1+x(1+x(1+x(1+x))))=x(1+x+x^2+x^3+x^4)$$

$$\frac{x^{(n-1)}(1+x+x^2+x^3+x^4)}{x(1+x+x^2+x^3+x^4)}=x^5$$

$$x^{(n-1)}=x^6$$
$$n-1=6$$
$$n=7$$

Ans: "B"
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Re: Tough Problem Solving Infinite Sequence Problem [#permalink]  13 Apr 2011, 12:25
I hope this is a 700+ level question...
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Re: Tough Problem Solving Infinite Sequence Problem [#permalink]  27 Dec 2011, 12:03
aznboi986 wrote:
I hope this is a 700+ level question...

this is 800+ level question
You definitely do not see any question like this one on real exam. GC problems usually too difficult.
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Sequence problem [#permalink]  18 Apr 2012, 18:36
In the infinite sequence A, the nth
term, A(n), is given by 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 A(n) to x(1+x(1+x
(1+x(1+x)))) equal to x^5?
(A) 8
(B) 7
(C) 6
(D) 5
(E) 4

Posted from my mobile device
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Re: Sequence problem [#permalink]  18 Apr 2012, 23:06
Expert's post
Ashikurrahman wrote:
In the infinite sequence A, the nth
term, A(n), is given by 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 A(n) to x(1+x(1+x
(1+x(1+x)))) equal to x^5?
(A) 8
(B) 7
(C) 6
(D) 5
(E) 4

Posted from my mobile device

Merging similar topics. Please ask if anything remains unclear.
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Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X [#permalink]  05 Jun 2012, 08:42
i) x(1 + x(1 + x(1 + x(1 + x)))) = x + x^2 + x^3 + x^4 + x^5
ii) An = x^(n-2) (x + x^2 + x^3 + x^4 + x^5)

Divide ii by i = x^(n-2)
We want ratio = x^5. So, n-2 = 5
=> n = 7. (B)
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Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X [#permalink]  24 Jul 2012, 18:27
omg.. by the time u read and digest the question its 1 minut e
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Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X [#permalink]  24 Jul 2012, 20:36
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rajathpanta wrote:
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]  30 Jul 2012, 21:28
Is this a typical 700 level GMAT question? or just an off topic question? experts pls advice.
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Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X [#permalink]  31 Jul 2012, 00:49
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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Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X [#permalink]  31 Jul 2012, 07:36
Expert's post
mohankumarbd wrote:
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]  31 Jul 2012, 08:35
Great Explanation +1
Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X   [#permalink] 31 Jul 2012, 08:35

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