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

It is currently 16 Aug 2018, 16:50

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

For every positive integer n, if the nth term of the sequence is xN

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

Hide Tags

Intern
Intern
avatar
Joined: 29 Jul 2016
Posts: 5
GMAT ToolKit User
For every positive integer n, if the nth term of the sequence is xN  [#permalink]

Show Tags

New post Updated on: 28 Sep 2016, 23:47
6
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

71% (02:06) correct 29% (01:41) wrong based on 66 sessions

HideShow timer Statistics

For every positive integer n, if the nth term of the sequence is \(x_n=\frac{1}{(n+1)(n+2)}\) then \(x_1+x_2+x_3+x_4+x_5=\)?

a) 4/5
b) 2/3
c) 5/14
d) 5/16
e) 6/19

Could anyone explain me how to do this without the normal convention of adding the 5 terms.

thanks in advance!

Originally posted by RidhimaNigam on 28 Sep 2016, 09:28.
Last edited by Bunuel on 28 Sep 2016, 23:47, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
Manager
Manager
avatar
Joined: 05 Jun 2015
Posts: 84
Location: United States
WE: Engineering (Transportation)
Re: For every positive integer n, if the nth term of the sequence is xN  [#permalink]

Show Tags

New post 28 Sep 2016, 18:32
2
2
Write \(x_n=\frac{1}{(n+1)(n+2)}\) as
\(x_n = \frac{1}{n+1} - \frac{1}{n+2}\)

\(x_1+x_2 = (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4})\)
= \(\frac{1}{2} - \frac{1}{4}\)

Only the first term and the last term remain when we add multiple \(x_n\) values.

\(x_5 = \frac{1}{6} - \frac{1}{7}\)
Therefore we have \(x_1+x_2+x_3+x_4+x_5= \frac{1}{2} - \frac{1}{7}\)
=\(\frac{5}{14}\)
Intern
Intern
avatar
Joined: 29 Jul 2016
Posts: 5
GMAT ToolKit User
Re: For every positive integer n, if the nth term of the sequence is xN  [#permalink]

Show Tags

New post 28 Sep 2016, 22:58
1
rudra316 wrote:
Write \(x_n=\frac{1}{(n+1)(n+2)}\) as
\(x_n = \frac{1}{n+1} - \frac{1}{n+2}\)

\(x_1+x_2 = (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4})\)
= \(\frac{1}{2} - \frac{1}{4}\)

Only the first term and the last term remain when we add multiple \(x_n\) values.

\(x_5 = \frac{1}{6} - \frac{1}{7}\)
Therefore we have \(x_1+x_2+x_3+x_4+x_5= \frac{1}{2} - \frac{1}{7}\)
=\(\frac{5}{14}\)

Oh yeah! Thanks

Sent from my ONE A2003 using GMAT Club Forum mobile app
Senior Manager
Senior Manager
User avatar
B
Status: Come! Fall in Love with Learning!
Joined: 05 Jan 2017
Posts: 433
Location: India
Premium Member
Re: For every positive integer n, if the nth term of the sequence is xN  [#permalink]

Show Tags

New post 27 Feb 2017, 06:56
1/(n+1)(n+2) can be written as 1/(n+1) - 1/(n+2)
Therefore x1 = ½ -⅓
x2 = ⅓ -¼ and so in.
Therefore x1+x2+x3+x4+x5 = ½-⅓ +⅓-¼ +¼ -⅕ +⅕ -⅙ +⅙ - 1/7 =½ - 1/7 = 5/14
_________________

GMAT Mentors
Image

Senior Manager
Senior Manager
User avatar
B
Status: Come! Fall in Love with Learning!
Joined: 05 Jan 2017
Posts: 433
Location: India
Premium Member
Re: For every positive integer n, if the nth term of the sequence is xN  [#permalink]

Show Tags

New post 27 Feb 2017, 06:56
1/(n+1)(n+2) can be written as 1/(n+1) - 1/(n+2)
Therefore x1 = ½ -⅓
x2 = ⅓ -¼ and so on.
Therefore x1+x2+x3+x4+x5 = ½-⅓ +⅓-¼ +¼ -⅕ +⅕ -⅙ +⅙ - 1/7 =½ - 1/7 = 5/14
_________________

GMAT Mentors
Image

Intern
Intern
avatar
B
Joined: 10 Feb 2018
Posts: 10
Re: For every positive integer n, if the nth term of the sequence is xN  [#permalink]

Show Tags

New post 05 Aug 2018, 17:43
RidhimaNigam wrote:
For every positive integer n, if the nth term of the sequence is \(x_n=\frac{1}{(n+1)(n+2)}\) then \(x_1+x_2+x_3+x_4+x_5=\)?

a) 4/5
b) 2/3
c) 5/14
d) 5/16
e) 6/19

Could anyone explain me how to do this without the normal convention of adding the 5 terms.

thanks in advance!


I must be missing something basic here, but what is the rule to rewrite the equation as below:

Write xn=1(n+1)(n+2) as
xn=1n+1−1n+2
Re: For every positive integer n, if the nth term of the sequence is xN &nbs [#permalink] 05 Aug 2018, 17:43
Display posts from previous: Sort by

For every positive integer n, if the nth term of the sequence is xN

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

Events & Promotions

PREV
NEXT


GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

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

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.