Last visit was: 21 Apr 2026, 19:16 It is currently 21 Apr 2026, 19:16
Home
Messages
Tests
Schools
Events
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
655-705 (Hard)|   Algebra|   Arithmetic|      
User avatar
liviofrol
User avatar
Joined: 30 Jan 2019
Last visit: 21 Aug 2024
Posts: 6
Own Kudos:
194
 [177]
Given Kudos: 29
Posts: 6
Kudos: 194
 [177]
Given that 1/n(n+1)) - 1/(n+1)(n+2) = 2/n(n+1)(n+2), which of the foll Updated on: 15 Dec 2023, 04:24
Originally posted by liviofrol on 12 Dec 2023, 15:09.
Last edited by Bunuel on 15 Dec 2023, 04:24, edited 1 time in total.
Edited the question.
6
Kudos
Add Kudos
169
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

55% (hard)

Question Stats:

75% (02:36) correct 25% (02:31) wrong based on 1038 sessions

History

Date
Time
Result
Not Attempted Yet
Given that \(\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} = \frac{2}{n(n+1)(n+2)}\), which of the following is equal to \(\frac{(2)(29)}{(5)(6)(7)}+\frac{(2)(29)}{(6)(7)(8)}+\frac{(2)(29)}{(7)(8)(9)}+ ... +\frac{(2)(29)}{(28)(29)(30)}\)?

​(A) \(\frac{14}{15}\)

(B) \(\frac{2}{35}\)

(C) \(\frac{117}{35}\)

(D) \(\frac{232}{35}\)

(E) \(\frac{232}{969}\)

Show Spoiler
Attachment:
Captura.PNG
Captura.PNG [ 24.31 KiB | Viewed 20216 times ]
ShowHide Answer
Official Answer
A
Trying to understand why the right answer is right? GMAT Club Tests explanations walk through concepts, shortcuts, and common traps on many questions. See how →
Most Helpful Reply
User avatar
umangshah
User avatar
Joined: 24 May 2023
Last visit: 18 Mar 2024
Posts: 67
Own Kudos:
97
 [35]
Given Kudos: 107
Status:Preparing for Exams
Location: India
GMAT 1: 640 Q44 V34
GPA: 4
WE:Consulting (Consulting)
GMAT 1: 640 Q44 V34
Posts: 67
Kudos: 97
 [35]
Given that 1/n(n+1)) - 1/(n+1)(n+2) = 2/n(n+1)(n+2), which of the foll 15 Dec 2023, 19:08
22
Kudos
Add Kudos
13
Bookmarks
Bookmark this Post
liviofrol
Given that \(\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} = \frac{2}{n(n+1)(n+2)}\), which of the following is equal to \(\frac{(2)(29)}{(5)(6)(7)}+\frac{(2)(29)}{(6)(7)(8)}+\frac{(2)(29)}{(7)(8)(9)}+ … +\frac{(2)(29)}{(28)(29)(30)}\)?

​(A) \(\frac{14}{15}\)

(B) \(\frac{2}{35}\)

(C) \(\frac{117}{35}\)

(D) \(\frac{232}{35}\)

(E) \(\frac{232}{969}\)

Show more

On expanding the equation using \(\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} = \frac{2}{n(n+1)(n+2)}\) , we can easily find a pattern.

Term 1: \(\frac{(2)(29)}{(5)(6)(7)}\)= \(\frac{29}{(5)(6)}-\frac{29}{(6)(7)}\)
Similarly term 2: \(\frac{(2)(29)}{(6)(7)(8)}\)=\(\frac{29}{(6)(7)}-\frac{29}{(7)(8)}\)

Pattern Continues upto \(\frac{(2)(29)}{(28)(29)(30)}\)

We can see that on adding term 1 and 2 \(\frac{29}{(6)(7)}\) cancels out. The same continues leaving us with only 1st and last term to consider.

Therefore, \(\frac{29}{(5)(6)}-\frac{29}{(29)(30)}= \frac{29(29-1)}{(29)(30)}=\frac{28}{30}=\frac{14}{15}\)
Option A is correct.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 21 Apr 2026
Posts: 109,728
Own Kudos:
810,476
 [10]
Given Kudos: 105,800
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,728
Kudos: 810,476
 [10]
Given that 1/n(n+1)) - 1/(n+1)(n+2) = 2/n(n+1)(n+2), which of the foll 21 Dec 2023, 23:53
2
Kudos
Add Kudos
8
Bookmarks
Bookmark this Post
AnkurGMAT20
Bunuel do you have similar questions for practice?
Show more

Try checking HERE.

Hope it helps.
General Discussion
User avatar
chetan2u
User avatar
GMAT Expert
Joined: 02 Aug 2009
Last visit: 18 Apr 2026
Posts: 11,230
Own Kudos:
44,984
 [5]
Given Kudos: 335
Status:Math and DI Expert
Location: India
Concentration: Human Resources, General Management
GMAT Focus 1: 735 Q90 V89 DI81
Products:
My Reviews
Expert
Expert reply
GMAT Focus 1: 735 Q90 V89 DI81
Posts: 11,230
Kudos: 44,984
 [5]
Given that 1/n(n+1)) - 1/(n+1)(n+2) = 2/n(n+1)(n+2), which of the foll 13 Dec 2023, 06:55
4
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
liviofrol
Which of the following is equal to \(\frac{(2)(29)}{(5)(6)(7)}+\frac{(2)(29)}{(6)(7)(8)}+\frac{(2)(29)}{(7)(8)(9)}+ … +\frac{(2)(29)}{(28)(29)(30)}\)?

​A) \(\frac{14}{15}\)

B) \(\frac{2}{35}\)

C) \(\frac{117}{35}\)

D) \(\frac{232}{35}\)

E) \(\frac{232}{969}\)
Show more


USE OPTIONS

Use options as they are wide spread.
\(\frac{(2)(29)}{(5)(6)(7)}+\frac{(2)(29)}{(6)(7)(8)}+\frac{(2)(29)}{(7)(8)(9)}+ … +\frac{(2)(29)}{(28)(29)(30)}\)
\(\frac{(2)(29)}{(5)(6)(7)}~=\frac{1}{4}......\frac{(2)(29)}{(6)(7)(8)}~=\frac{1}{6}......\frac{(2)(29)}{(7)(8)(9)}~=\frac{1}{9}...... … \frac{(2)(29)}{(28)(29)(30)}~=\frac{1}{420}......\)

(1) The value has to be greater than 1/3, so discard B and E
(2) The first value is 1/3, then it reduces drastically with each successive term...1/3, 1/4, 1/9, all other terms are less than 1/10, and it reaches 1/140
Clearly, the sum will around 1 or at the max 1.5. Only possibility 14/15
(3) For the answer to be 117/35 or nearly 3 for the sum of 24 terms, each term should be nearly 3/24 or 1/8. But only 2 terms are greater than 1/8.

Approximation
Take 29 as 30 and the actual answer answer should be lesser than what you get.
\(\frac{(2)(30)}{(5)(6)(7)}+\frac{(2)(30)}{(6)(7)(8)}+\frac{(2)(30)}{(7)(8)(9)}+ … +\frac{(2)(30)}{(28)(29)(30)}\)
\(\frac{10}{(5)(7)}+\frac{10}{(7)(8)}+\frac{10}{(7)(4)(3)}+ … +\frac{10}{(14)(30)}\)
\(\frac{1}{3.5}+\frac{1}{5.6}+\frac{1}{8.4}+\frac{1}{12}+\frac{1}{16} … +\frac{1}{(140)}\)
All options can be discarded except 14/15

Proper method

\(\frac{(2)(29)}{(5)(6)(7)}+\frac{(2)(29)}{(6)(7)(8)}+\frac{(2)(29)}{(7)(8)(9)}+ … +\frac{(2)(29)}{(28)(29)(30)}\)

Now, \(\frac{1}{(5)(6)(7)}=\frac{1}{2*5}-\frac{1}{6}+\frac{1}{2*7}............\frac{1}{(6)(7)(8)}=\frac{1}{2*6}-\frac{1}{7}+\frac{1}{2*8}............\frac{(2)(29)}{(7)(8)(9)}=\frac{1}{2*7}-\frac{1}{8}+\frac{1}{2*9}............\)

\(\frac{(2)(29)}{(5)(6)(7)}+\frac{(2)(29)}{(6)(7)(8)}+\frac{(2)(29)}{(7)(8)(9)}+ … +\frac{(2)(29)}{(28)(29)(30)}\)
\(2*29(\frac{1}{2*5}-\frac{1}{6}+\frac{1}{2*7}+\frac{1}{2*6}-\frac{1}{7}+\frac{1}{2*8}+\frac{1}{2*7}-\frac{1}{8}+\frac{1}{2*9}............\)

Terms will get cancelled out => \(\frac{1}{2*7}-\frac{1}{7}+\frac{1}{2*7}=0\)
\(2*29(\frac{1}{2*5}-\frac{1}{2*6}-\frac{1}{2*29}+\frac{1}{2*30})\)
\(29(\frac{1}{5}-\frac{1}{6}-\frac{1}{29}+\frac{1}{30})\)
\(29(\frac{1}{5}-\frac{1}{6}+\frac{1}{30})-\frac{29}{29}\)
\(29(\frac{6-5+1}{30})-\frac{29}{29}=\frac{2*29}{30}-1=\frac{28}{30}=\frac{14}{15}\)
User avatar
AnkurGMAT20
User avatar
Joined: 12 Jul 2020
Last visit: 13 Mar 2025
Posts: 134
Own Kudos:
1,417
 [2]
Given Kudos: 263
Location: India
Schools: IIM IIM IIM IIM
GMAT Focus 1: 615 Q79 V83 DI79
GMAT 1: 530 Q42 V20
GMAT 2: 640 Q48 V29
GPA: 6.23
Schools: IIM IIM IIM IIM
GMAT Focus 1: 615 Q79 V83 DI79
GMAT 2: 640 Q48 V29
Posts: 134
Kudos: 1,417
 [2]
Given that 1/n(n+1)) - 1/(n+1)(n+2) = 2/n(n+1)(n+2), which of the foll 21 Dec 2023, 22:21
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel do you have similar questions for practice?
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 21 Apr 2026
Posts: 16,438
Own Kudos:
79,375
 [5]
Given Kudos: 484
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,438
Kudos: 79,375
 [5]
Given that 1/n(n+1)) - 1/(n+1)(n+2) = 2/n(n+1)(n+2), which of the foll 11 Jan 2025, 00:13
5
Kudos
Add Kudos
Bookmarks
Bookmark this Post
liviofrol
Given that \(\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} = \frac{2}{n(n+1)(n+2)}\), which of the following is equal to \(\frac{(2)(29)}{(5)(6)(7)}+\frac{(2)(29)}{(6)(7)(8)}+\frac{(2)(29)}{(7)(8)(9)}+ ... +\frac{(2)(29)}{(28)(29)(30)}\)?

(A) \(\frac{14}{15}\)

(B) \(\frac{2}{35}\)

(C) \(\frac{117}{35}\)

(D) \(\frac{232}{35}\)

(E) \(\frac{232}{969}\)

Show Spoiler
Attachment:
Captura.PNG
Show more

It is a Series question in which they have given you how to break down the term to make the question easier. Given:

\(\frac{(2)(29)}{(5)(6)(7)}+\frac{(2)(29)}{(6)(7)(8)}+\frac{(2)(29)}{(7)(8)(9)}+ ... +\frac{(2)(29)}{(28)(29)(30)}\)

Take 29 common:

\(29 * [\frac{(2)}{(5)(6)(7)}+\frac{(2)}{(6)(7)(8)}+\frac{(2)}{(7)(8)(9)}+ ... +\frac{(2)}{(28)(29)(30)}]\)

Each of these terms can be broken down as per the given relation:

\(\frac{(2)}{(5)(6)(7)} = \frac{1}{(5)(6)} - \frac{1}{(6)(7)}\)

\(\frac{(2)}{(6)(7)(8)} = \frac{1}{(6)(7)} - \frac{1}{(7)(8)}\)

Note the pattern. When we add them all, 1/6*7 will get cancelled off. Similarly, all terms will get cancelled off except the first and last terms. We will get

\(29 * [\frac{1}{(5)(6)} - \frac{1}{(29)(30)}] = 29 * \frac{29 - 1}{29*30} = \frac{14}{15}\)

Answer (A)

Here is a discussion on how to solve Series Questions: https://youtu.be/KX8WNiyNUIo
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,957
Own Kudos:
1,117
Posts: 38,957
Kudos: 1,117
Given that 1/n(n+1)) - 1/(n+1)(n+2) = 2/n(n+1)(n+2), which of the foll 12 Jan 2026, 11:58
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.
Moderators:
Bunuel
Math Expert
109728 posts
Krunaal
Tuck School Moderator
853 posts