Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 27 Oct 2016, 04:59

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Intern
Joined: 31 Oct 2010
Posts: 32
Followers: 0

Kudos [?]: 57 [2] , given: 25

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

### Show Tags

13 Dec 2010, 07:59
2
KUDOS
8
This post was
BOOKMARKED
00:00

Difficulty:

75% (hard)

Question Stats:

62% (03:48) correct 38% (02:04) wrong based on 213 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?

(1) j>8
(2) j<16
[Reveal] Spoiler: OA
Intern
Joined: 31 Oct 2010
Posts: 32
Followers: 0

Kudos [?]: 57 [0], given: 25

Re: Knewton Diagnostic Question #28 [#permalink]

### Show Tags

13 Dec 2010, 08:09
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
Math Expert
Joined: 02 Sep 2009
Posts: 35317
Followers: 6648

Kudos [?]: 85826 [4] , given: 10254

Re: Knewton Diagnostic Question #28 [#permalink]

### Show Tags

13 Dec 2010, 08:24
4
KUDOS
Expert's post
4
This post was
BOOKMARKED
mmcooley33 wrote:
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:

$$t_1=\frac{1}{1^2}-\frac{1}{(1+1)^2}=1-\frac{1}{2^2}$$;

$$t_2=\frac{1}{2^2}-\frac{1}{(2+1)^2}=\frac{1}{2^2}-\frac{1}{3^2}$$;

$$t_3=\frac{1}{3^2}-\frac{1}{(3+1)^2}=\frac{1}{3^2}-\frac{1}{4^2}$$;
...

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.

_________________
Senior Manager
Status: Bring the Rain
Joined: 17 Aug 2010
Posts: 406
Location: United States (MD)
Concentration: Strategy, Marketing
Schools: Michigan (Ross) - Class of 2014
GMAT 1: 730 Q49 V39
GPA: 3.13
WE: Corporate Finance (Aerospace and Defense)
Followers: 7

Kudos [?]: 44 [0], given: 46

Re: Knewton Diagnostic Question #28 [#permalink]

### Show Tags

13 Dec 2010, 08:36
These pattern questions always get me.

Thanks for the explanation
_________________
Manager
Status: Waiting
Joined: 11 Dec 2012
Posts: 53
Location: Bahrain
Concentration: Healthcare, General Management
GMAT 1: 640 Q49 V24
GMAT 2: 720 Q49 V40
WE: Sales (Other)
Followers: 6

Kudos [?]: 56 [4] , given: 144

Re: Knewton Diagnostic Question #28 [#permalink]

### Show Tags

02 Apr 2013, 10:56
4
KUDOS
Bunuel wrote:

Question: is $$Sum_j=1-\frac{1}{(j+1)^2}>\frac{63}{64}$$ --> is $$(j+1)^2>64$$--> is $$j>8$$?

Dear Bunuel
Should it not be
$$(j+1)^2>64$$--> is $$j>7$$

Bunuel wrote:
]--> is $$j>8$$?

Thank you
Math Expert
Joined: 02 Sep 2009
Posts: 35317
Followers: 6648

Kudos [?]: 85826 [0], given: 10254

Re: Knewton Diagnostic Question #28 [#permalink]

### Show Tags

02 Apr 2013, 11:00
jainpiyushjain wrote:
Bunuel wrote:

Question: is $$Sum_j=1-\frac{1}{(j+1)^2}>\frac{63}{64}$$ --> is $$(j+1)^2>64$$--> is $$j>8$$?

Dear Bunuel
Should it not be
$$(j+1)^2>64$$--> is $$j>7$$

Bunuel wrote:
]--> is $$j>8$$?

Thank you

Typo edited. Thank you. +1.
_________________
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 6972
Location: Pune, India
Followers: 2033

Kudos [?]: 12782 [1] , given: 221

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

### Show Tags

05 Feb 2014, 07:38
1
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.

_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Intern
Joined: 04 Jun 2014
Posts: 49
Followers: 0

Kudos [?]: 2 [0], given: 5

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

### Show Tags

13 Sep 2014, 06:54
Can someone explain what the difference between this question here and this question is: the-sequence-s1-s2-s3-sn-is-such-that-sn-1-n-103947.html

Seems like they kinda ask the same, but I dont understand why the final question here is:
is (j+1)^2>64--> is j>7 ?

And in the other question we ask about the numerator, is k > 9 ?
Is there any difference?
Intern
Joined: 29 Aug 2013
Posts: 5
Followers: 0

Kudos [?]: 0 [0], given: 1

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

### Show Tags

16 Sep 2014, 20:56
Bunuel wrote:
jainpiyushjain wrote:
Bunuel wrote:

Question: is $$Sum_j=1-\frac{1}{(j+1)^2}>\frac{63}{64}$$ --> is $$(j+1)^2>64$$--> is $$j>8$$?

Dear Bunuel
Should it not be
$$(j+1)^2>64$$--> is $$j>7$$

Bunuel wrote:
]--> is $$j>8$$?

Thank you

Typo edited. Thank you. +1.

I think j>7 is correct since J+1>8....So J>7, isn't it?
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 12253
Followers: 542

Kudos [?]: 151 [0], given: 0

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.
_________________
Re: The numbers above form a sequence, t1, t2, and t3 , which is   [#permalink] 30 Nov 2015, 04:37
Similar topics Replies Last post
Similar
Topics:
t1, t2, t3, t4...tn what is the value of n? 1) Sum of n 8 09 Jan 2007, 07:43
5 Is t^2 > t/3? 11 12 Mar 2010, 23:52
10 If the terms of a sequence are t1, t2, t3, . . . , tn, what 9 10 Jan 2009, 23:52
2 In the sequence of non-zero numbers t1, t2,t3,......tn,..... 4 12 Jul 2008, 22:34
17 In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1 15 18 Jun 2008, 10:23
Display posts from previous: Sort by