Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 1612:00 PM EDT
-11:59 PM EDT
You’re first to know: Save $200 on GMAT prep starting tomorrow, 4/13. Get 6 official practice tests and study with real questions. Be ready to save!
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
13 Dec 2010, 07:59
Kudos
Bookmarks
A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
61% (02:41) correct
39%
(02:28)
wrong
based on 494
sessions
History
Date
Time
Result
Not Attempted Yet
3/4, 5/36, 7/144
The numbers above form a sequence, t1, t2, and t3 , which is defined by \(t_m=\frac{1}{m^2}-\frac{1}{(m+1)^2}\) for all positive integers m. Is the sum of the first J terms of the sequence greater than 63/64?
(1) j>8
(2) j<16
The numbers above form a sequence, t1, t2, and t3 , which is defined by \(t_m=\frac{1}{m^2}-\frac{1}{(m+1)^2}\) for all positive integers m. Is the sum of the first J terms of the sequence greater than 63/64?
(1) j>8
(2) j<16
13 Dec 2010, 08:24
Kudos
Bookmarks
mmcooley33
In such kind of questions there is always a pattern in terms or/and in the sum of the terms.
Given: \(t_m=\frac{1}{m^2}-\frac{1}{(m+1)^2}\). So:
\(t_1=\frac{1}{1^2}-\frac{1}{(1+1)^2}=1-\frac{1}{2^2}\);
\(t_2=\frac{1}{2^2}-\frac{1}{(2+1)^2}=\frac{1}{2^2}-\frac{1}{3^2}\);
\(t_3=\frac{1}{3^2}-\frac{1}{(3+1)^2}=\frac{1}{3^2}-\frac{1}{4^2}\);
...
You should notice that if we have as sum of first 2 terms then every thing but the 1 from \(t_1\) and the last part from \(t_2\) (1/3^2=1/(2+1)^2) will cancel out, so \(sum_2=1-\frac{1}{(2+1)^2}\). The same if we sum first 3 terms: only 1 minus the last part of \(t_3\) (1/4^2=1/(3+1)^2) will remain, \(sum_3=1-\frac{1}{(3+1)^2}\). So if we sum first \(j\) terms the the sum will equal to \(1-\frac{1}{(j+1)^2}\).
Question: is \(Sum_j=1-\frac{1}{(j+1)^2}>\frac{63}{64}\) --> is \((j+1)^2>64\)--> is \(j>7\)?
(1) j>8. Sufficient.
(2) j<16. Not sufficient.
Answer: A.
General Discussion
13 Dec 2010, 08:09
Kudos
Bookmarks
the solution has to do with manipulation of the negative. The answer given by knewton turns 1/m^2 - 1/((m+1)^2) into (-1/M^2 + 1/m^2) so i get how it turns all but 1 into zero but they say the last term (-1/j^2 + 1/j^2) - 1/((j+1)^2). I am probably missing a simple well known concept from order of operations but could someone please enlighten me. Plus if there is a better way to cut and paste formulas from websites please inform me thanks, otherwise i would post their solution for reference. Thanks a bunch
