GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 07 Dec 2019, 16:42

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Intern
Intern
avatar
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]

Show Tags

New post 13 Sep 2014, 09:18
FYI guys, this is a Manhattan Review Turbocharge question.
Manager
Manager
avatar
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]

Show Tags

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..
Intern
Intern
avatar
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]

Show Tags

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
Intern
Intern
avatar
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]

Show Tags

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.
Manager
Manager
avatar
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]

Show Tags

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 882 times ]

Senior Manager
Senior Manager
User avatar
P
Joined: 29 Jun 2017
Posts: 417
GPA: 4
WE: Engineering (Transportation)
GMAT ToolKit User Reviews Badge
Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X^(n+2) + X  [#permalink]

Show Tags

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
Manager
Manager
User avatar
B
Joined: 24 Jan 2017
Posts: 66
Location: Brazil
Concentration: Strategy, Entrepreneurship
GPA: 3.5
WE: Consulting (Consulting)
CAT Tests
Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X^(n+2) + X  [#permalink]

Show Tags

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!!!
Intern
Intern
User avatar
B
Joined: 07 Jun 2018
Posts: 20
Location: India
Concentration: Marketing, International Business
GPA: 3.9
WE: Marketing (Insurance)
GMAT ToolKit User
Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X^(n+2) + X  [#permalink]

Show Tags

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.
Veritas Prep GMAT Instructor
User avatar
V
Joined: 16 Oct 2010
Posts: 9850
Location: Pune, India
Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X^(n+2) + X  [#permalink]

Show Tags

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))))
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 13724
Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X^(n+2) + X  [#permalink]

Show Tags

New post 21 Aug 2019, 08:35
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Club Bot
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

Go to page   Previous    1   2   [ 30 posts ] 

Display posts from previous: Sort by

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne