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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

The numbers above form a sequence, t1, t2, and t3 , which is [#permalink]

Show Tags

13 Dec 2010, 07:59

2

This post received KUDOS

8

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

75% (hard)

Question Stats:

61% (03:50) correct
39% (02:04) wrong based on 205 sessions

HideShow timer Statistics

3/4, 5/36, 7/144 The numbers above form a sequence, t1, t2, and t3 , which is defined by tm = (1 / m^2) - (1/m+1^2)for all positive integers m. Is the sum of the first J terms of the sequence greater than 63/64?

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

3/4, 5/36, 7/144 The numbers above form a sequence,t1,t2, and t3 , which is defined by tm = (1 / m^2) - (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

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:

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 \(j>7\)?

The numbers above form a sequence, t1, t2, and t3 [#permalink]

Show Tags

05 Feb 2014, 07:38

1

This post received KUDOS

Expert's post

Responding to a pm:

Question: 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

Solution: The numbers given above don't help us in any calculations. We should try to write the sequence on our own. \(t_m = \frac{1}{m^2} - \frac{1}{(m+1)^2}\) \(t_1 = 1 - 1/4\) \(t_2 = 1/4 - 1/9\) \(t_3 = 1/9 - 1/16\)

Notice that the second term cancels out the first term of the next number. So when we add all these numbers, we will be left with 1 - the second term of the last number (because it will not get canceled)

(1) j>8

The number of terms will be at least 9. The sum of first 9 terms \(= 1 - \frac{1}{10^2} = 99/100\) This is greater than 63/64. As the number of terms keep increasing, the second term which is subtracted keeps getting smaller so the sum tends toward 1. Hence the sum will always be greater than 63/64. Sufficient.

(2) j<16 The number of terms could be 1 or 9 or 15 etc If the number of terms is 1, the sum will be 3/4 which is less than 63/64. As discussed in statement 1, if the number of terms is 9, the sum will be greater than 63/64. Not sufficient.

Re: The numbers above form a sequence, t1, t2, and t3 , which is [#permalink]

Show Tags

30 Nov 2015, 04:37

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. _________________

Excellent posts dLo saw your blog too..!! Man .. you have got some writing skills. And Just to make an argument = You had such an amazing resume ; i am glad...

So Much $$$ Business school costs a lot. This is obvious, whether you are a full-ride scholarship student or are paying fully out-of-pocket. Aside from the (constantly rising)...