In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X^(n+2) + X

Question

In the infinite sequence A,  A_n = 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

Bunuel · Accepted Answer

In the infinite sequence A,  A_n = 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

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)

--> work on the right side expression:

-->  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

Answer: B.

craky · Answer

Its not so hard when you realize how can you solve it, but until that, you spent half of your life.

KarishmaB · Answer

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.

ksharma12 · Answer

great explanation. Thanks. +1 to you

subhashghosh · Answer

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

KarishmaB · Answer

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

fluke · Answer

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"

rajathpanta · Answer

omg.. by the time u read and digest the question its 1 minut :roll:

e

KarishmaB · Answer

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.

EvaJager · Answer

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...

KarishmaB · Answer

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.

Gmbrox · Answer

I did it like it : as you can see x(1+x(1+x(1+x(1+x)))) X comes 5 times, therefore the max term will be X^5, in the question you see that you want to arrive at X^5 so it means that in the sum of X^n-1...x^n+3 the max term must be X^10 so that it can be x^5(x^5) therefore 10 = 3+n, n=7, timer indicate me 1min 53.

But definitely i had the answer, but i was unable to demonstrate it in that time, it would take more like 5 to 10 minutes.

But definitely i had the answer, but i was unable to demonstrate it in that time, it would take more like 5 to 10 minutes.

knicks1288 · Answer

Could someone please explain to me how that's an infinite sequence? That's what really threw me off.

KarishmaB · Answer

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.

Will2020 · Answer

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!!!

KarishmaB · Answer

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) )))

Bambi2021 · Answer

Number plugging actually was a good idea here. n = 4 gives (n-1) = 3, so the least exponent in the expression will be 3.

Now we know that x^3 is the highest term common for all factors of the expression that we can pull out of the expression to arrive at: x^3(1 + x + x^2...)

This reminds a lot of what we are going to divide with.

x^3(1 + x + x^2...) divided with
x(1 + x(1 + x(...)

Let's just guess that the brackets equal out and we are left with:

x^3 / x

The ratio here is x^2 so we need to increase the exponent by 3 to get x^6. And what does the 6 represent? Thats right, n-1.

n = 7

gmatophobia · Answer

A_n = x^{(n-1)} + x^n + x^{(n+1)} + x^{(n+2)} + x^{(n+3)}

A_n = x^{n-1}( 1 + x + x^{2} + x^{3} + x^{4})

Assume x = 2 (We can assume a value so as to reduce the expression)

A_n = 2^{n-1}(1 + 2 + 4 + 8 + 16) = 2^{n-1}(31)

Value of  x(1+x(1+x(1+x(1+x))))  at x = 2

2(1+2(1+2(1+2(1+2))))  = 2*31

It's given that

\frac{2^{n-1}(31) }{ 2*31} = 2^{5}

2^{n-2} = 2^{5}

n-2 = 5

n = 7

Option B

krishna_sunder_ · Answer

take x as something and solve guys...
i took x = 2.......don't make the mistake of taking x = 1, as 1 to the power of anything is 1...so finding n like this will become difficult or maybe impossible...

once u take x=2, 
first equation/ second equation = 2^5
now it's very simple from here....
because, as u substitute x=2 to the second equation...u get a value....
and solving for n becomes a piece of cake

In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X^(n+2) + X

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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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