For all positive integers n, the sequence An is defined by : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 20 Jan 2017, 23:11

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

For all positive integers n, the sequence An is defined by

Author Message
TAGS:

Hide Tags

Manager
Status: Never ever give up on yourself.Period.
Joined: 23 Aug 2012
Posts: 152
Location: India
Concentration: Finance, Human Resources
GMAT 1: 570 Q47 V21
GMAT 2: 690 Q50 V33
GPA: 3.5
WE: Information Technology (Investment Banking)
Followers: 10

Kudos [?]: 303 [3] , given: 35

For all positive integers n, the sequence An is defined by [#permalink]

Show Tags

22 Dec 2012, 05:34
3
KUDOS
9
This post was
BOOKMARKED
00:00

Difficulty:

85% (hard)

Question Stats:

50% (03:23) correct 50% (02:27) wrong based on 175 sessions

HideShow timer Statistics

For all positive integers n, the sequence $$A_n$$ is defined by the following relationship:

$$A_n = \frac{n-1}{n!}$$

What is the sum of all the terms in the sequence from $$A_1$$ through$$A_{10}$$, inclusive?

(A) $$\frac{9!+1}{10!}$$

(B) $$\frac{9(9!)}{10!}$$

(C) $$\frac{10!-1}{10!}$$

(D) $$\frac{10!}{10!+1}$$

(E) $$\frac{10(10!)}{11!}$$
[Reveal] Spoiler: OA

_________________

Don't give up on yourself ever. Period.
Beat it, no one wants to be defeated (My journey from 570 to 690) : http://gmatclub.com/forum/beat-it-no-one-wants-to-be-defeated-journey-570-to-149968.html

Math Expert
Joined: 02 Sep 2009
Posts: 36583
Followers: 7087

Kudos [?]: 93294 [9] , given: 10555

Re: For all positive integers n, the sequence An is defined by [#permalink]

Show Tags

22 Dec 2012, 05:49
9
KUDOS
Expert's post
2
This post was
BOOKMARKED
For all positive integers n, the sequence $$A_n$$ is defined by the following relationship:

$$A_n = \frac{n-1}{n!}$$

What is the sum of all the terms in the sequence from $$A_1$$ through$$A_{10}$$, inclusive?

(A) $$\frac{9!+1}{10!}$$

(B) $$\frac{9(9!)}{10!}$$

(C) $$\frac{10!-1}{10!}$$

(D) $$\frac{10!}{10!+1}$$

(E) $$\frac{10(10!)}{11!}$$
_____________________________________

$$A_1=0$$;
$$A_2=\frac{1}{2!}$$;
$$A_3=\frac{2}{3!}$$;
$$A_4=\frac{3}{4!}$$;
...
$$A_{10}=\frac{9}{10!}$$.

The sum of the first 2 terms is: $$A_1+A_2=\frac{1}{2!}=\frac{2!-1}{2!}$$;
The sum of the first 3 terms is: $$A_1+A_2+A_3=\frac{5}{3!}=\frac{3!-1}{3!}$$;
The sum of the first 4 terms is: $$A_1+A_2+A_3+A_4=\frac{23}{4!}=\frac{4!-1}{4!}$$.

Similarly the sum of the first 10 terms is $$\frac{10!-1}{10!}$$.

_________________
Manager
Joined: 26 Sep 2013
Posts: 221
Concentration: Finance, Economics
GMAT 1: 670 Q39 V41
GMAT 2: 730 Q49 V41
Followers: 4

Kudos [?]: 139 [0], given: 40

Re: For all positive integers n, the sequence An is defined by [#permalink]

Show Tags

24 Nov 2013, 12:55
Bunuel wrote:
For all positive integers n, the sequence $$A_n$$ is defined by the following relationship:

$$A_n = \frac{n-1}{n!}$$

What is the sum of all the terms in the sequence from $$A_1$$ through$$A_{10}$$, inclusive?

(A) $$\frac{9!+1}{10!}$$

(B) $$\frac{9(9!)}{10!}$$

(C) $$\frac{10!-1}{10!}$$

(D) $$\frac{10!}{10!+1}$$

(E) $$\frac{10(10!)}{11!}$$
_____________________________________

$$A_1=0$$;
$$A_2=\frac{1}{2!}$$;
$$A_3=\frac{2}{3!}$$;
$$A_4=\frac{3}{4!}$$;
...
$$A_{10}=\frac{9}{10!}$$.

The sum of the first 2 terms is: $$A_1+A_2=\frac{1}{2!}=\frac{2!-1}{2!}$$;
The sum of the first 3 terms is: $$A_1+A_2+A_3=\frac{5}{3!}=\frac{3!-1}{3!}$$;
The sum of the first 4 terms is: $$A_1+A_2+A_3+A_4=\frac{23}{4!}=\frac{4!-1}{4!}$$.

Similarly the sum of the first 10 terms is $$\frac{10!-1}{10!}$$.

I tried a creative solution that didn't work, and was hoping you could shed light on why.

$$(\frac{(A2+A10)}{2})*9$$

(since A1 is irrelevant, as it is 0, only A2-A10 are needed for the total)

Ended up with $$((\frac{1}{2!}+\frac{9}{10!})/2)*9$$, which clearly is not going to lend a correct answer. What was wrong with using the formula to find the sum of a sequence, in this case?
Math Expert
Joined: 02 Sep 2009
Posts: 36583
Followers: 7087

Kudos [?]: 93294 [0], given: 10555

Re: For all positive integers n, the sequence An is defined by [#permalink]

Show Tags

24 Nov 2013, 13:04
AccipiterQ wrote:
Bunuel wrote:
For all positive integers n, the sequence $$A_n$$ is defined by the following relationship:

$$A_n = \frac{n-1}{n!}$$

What is the sum of all the terms in the sequence from $$A_1$$ through$$A_{10}$$, inclusive?

(A) $$\frac{9!+1}{10!}$$

(B) $$\frac{9(9!)}{10!}$$

(C) $$\frac{10!-1}{10!}$$

(D) $$\frac{10!}{10!+1}$$

(E) $$\frac{10(10!)}{11!}$$
_____________________________________

$$A_1=0$$;
$$A_2=\frac{1}{2!}$$;
$$A_3=\frac{2}{3!}$$;
$$A_4=\frac{3}{4!}$$;
...
$$A_{10}=\frac{9}{10!}$$.

The sum of the first 2 terms is: $$A_1+A_2=\frac{1}{2!}=\frac{2!-1}{2!}$$;
The sum of the first 3 terms is: $$A_1+A_2+A_3=\frac{5}{3!}=\frac{3!-1}{3!}$$;
The sum of the first 4 terms is: $$A_1+A_2+A_3+A_4=\frac{23}{4!}=\frac{4!-1}{4!}$$.

Similarly the sum of the first 10 terms is $$\frac{10!-1}{10!}$$.

I tried a creative solution that didn't work, and was hoping you could shed light on why.

$$(\frac{(A2+A10)}{2})*9$$

(since A1 is irrelevant, as it is 0, only A2-A10 are needed for the total)

Ended up with $$((\frac{1}{2!}+\frac{9}{10!})/2)*9$$, which clearly is not going to lend a correct answer. What was wrong with using the formula to find the sum of a sequence, in this case?

I guess you wanted to find the sum with the formula for evenly spaced sets: (first+last)/2*(# of terms). But the given sequence is not evenly spaced, thus you cannot apply it here.

Hope it's clear.
_________________
Manager
Joined: 26 Sep 2013
Posts: 221
Concentration: Finance, Economics
GMAT 1: 670 Q39 V41
GMAT 2: 730 Q49 V41
Followers: 4

Kudos [?]: 139 [0], given: 40

Re: For all positive integers n, the sequence An is defined by [#permalink]

Show Tags

24 Nov 2013, 13:23
Bunuel wrote:
AccipiterQ wrote:
Bunuel wrote:
For all positive integers n, the sequence $$A_n$$ is defined by the following relationship:

$$A_n = \frac{n-1}{n!}$$

What is the sum of all the terms in the sequence from $$A_1$$ through$$A_{10}$$, inclusive?

(A) $$\frac{9!+1}{10!}$$

(B) $$\frac{9(9!)}{10!}$$

(C) $$\frac{10!-1}{10!}$$

(D) $$\frac{10!}{10!+1}$$

(E) $$\frac{10(10!)}{11!}$$
_____________________________________

$$A_1=0$$;
$$A_2=\frac{1}{2!}$$;
$$A_3=\frac{2}{3!}$$;
$$A_4=\frac{3}{4!}$$;
...
$$A_{10}=\frac{9}{10!}$$.

The sum of the first 2 terms is: $$A_1+A_2=\frac{1}{2!}=\frac{2!-1}{2!}$$;
The sum of the first 3 terms is: $$A_1+A_2+A_3=\frac{5}{3!}=\frac{3!-1}{3!}$$;
The sum of the first 4 terms is: $$A_1+A_2+A_3+A_4=\frac{23}{4!}=\frac{4!-1}{4!}$$.

Similarly the sum of the first 10 terms is $$\frac{10!-1}{10!}$$.

I tried a creative solution that didn't work, and was hoping you could shed light on why.

$$(\frac{(A2+A10)}{2})*9$$

(since A1 is irrelevant, as it is 0, only A2-A10 are needed for the total)

Ended up with $$((\frac{1}{2!}+\frac{9}{10!})/2)*9$$, which clearly is not going to lend a correct answer. What was wrong with using the formula to find the sum of a sequence, in this case?

I guess you wanted to find the sum with the formula for evenly spaced sets: (first+last)/2*(# of terms). But the given sequence is not evenly spaced, thus you cannot apply it here.

Hope it's clear.

oh my goodness, I completely overlooked that in my zeal to be clever
Current Student
Joined: 03 Feb 2013
Posts: 939
Location: India
Concentration: Operations, Strategy
GMAT 1: 760 Q49 V44
GPA: 3.88
WE: Engineering (Computer Software)
Followers: 134

Kudos [?]: 844 [0], given: 546

Re: For all positive integers n, the sequence An is defined by [#permalink]

Show Tags

20 Jan 2014, 08:26
daviesj wrote:
For all positive integers n, the sequence $$A_n$$ is defined by the following relationship:

$$A_n = \frac{n-1}{n!}$$

What is the sum of all the terms in the sequence from $$A_1$$ through$$A_{10}$$, inclusive?

(A) $$\frac{9!+1}{10!}$$

(B) $$\frac{9(9!)}{10!}$$

(C) $$\frac{10!-1}{10!}$$

(D) $$\frac{10!}{10!+1}$$

(E) $$\frac{10(10!)}{11!}$$

One of radical theory to solve this kind of sequence problem is

Every problem which has such series, is also true for n, so this value will be true for value of n

A1 = 0, A2 = 1/2 , A3 = 1/3

Sum of A1 - A3 = 1/2 + 1/3 = 5/6
If the above sequence is correct, we can take one option after another and just substitute n = 2.
For example
$$\frac{9!+1}{10!}$$ has to be true for n and nth term will be $$\frac{(n-1)!+1}{n!}$$
So we can put n = 3 and verify.
Hence the option A) 3/6 = 1/2 Eliminated.
Option B) $$\frac{9(9!)}{10!}$$ - In terms of n - $$\frac{(n-1)(n-1)!}{n!}$$
2*2/6 = 2/3 Eliminated.
Option C) $$\frac{10!-1}{10!}$$ - In terms of n - $$\frac{n!-1}{n!}$$
5/6 - Hold
Option D) $$\frac{10!}{10!+1}$$ - In terms of n - $$\frac{n!}{n!+1}$$
6/7 - Eliminated
Option e) $$\frac{10(10!)}{11!}$$[/quote] - in terms of n - $$\frac{n(n!)}{(n+1)!}$$
Without even substituting, Eliminated !!!

Option C)

My Sir used to say, in this type of questions marking takes more time..
_________________

Thanks,
Kinjal

My Application Experience : http://gmatclub.com/forum/hardwork-never-gets-unrewarded-for-ever-189267-40.html#p1516961
Prodigy for Tepper - CMU : http://bit.ly/cmuloan-kd

Intern
Status: Yes, I can and I will. 700+FTW
Joined: 30 Sep 2013
Posts: 20
Location: India
Concentration: Finance, Strategy
GMAT Date: 02-05-2014
GPA: 3.7
WE: Other (Retail)
Followers: 2

Kudos [?]: 15 [0], given: 3

Re: For all positive integers n, the sequence An is defined by [#permalink]

Show Tags

24 Jan 2014, 02:31
are there more such type of questions bunnel?

Bunuel wrote:
AccipiterQ wrote:
Bunuel wrote:
For all positive integers n, the sequence $$A_n$$ is defined by the following relationship:

$$A_n = \frac{n-1}{n!}$$

What is the sum of all the terms in the sequence from $$A_1$$ through$$A_{10}$$, inclusive?

(A) $$\frac{9!+1}{10!}$$

(B) $$\frac{9(9!)}{10!}$$

(C) $$\frac{10!-1}{10!}$$

(D) $$\frac{10!}{10!+1}$$

(E) $$\frac{10(10!)}{11!}$$
_____________________________________

$$A_1=0$$;
$$A_2=\frac{1}{2!}$$;
$$A_3=\frac{2}{3!}$$;
$$A_4=\frac{3}{4!}$$;
...
$$A_{10}=\frac{9}{10!}$$.

The sum of the first 2 terms is: $$A_1+A_2=\frac{1}{2!}=\frac{2!-1}{2!}$$;
The sum of the first 3 terms is: $$A_1+A_2+A_3=\frac{5}{3!}=\frac{3!-1}{3!}$$;
The sum of the first 4 terms is: $$A_1+A_2+A_3+A_4=\frac{23}{4!}=\frac{4!-1}{4!}$$.

Similarly the sum of the first 10 terms is $$\frac{10!-1}{10!}$$.

I tried a creative solution that didn't work, and was hoping you could shed light on why.

$$(\frac{(A2+A10)}{2})*9$$

(since A1 is irrelevant, as it is 0, only A2-A10 are needed for the total)

Ended up with $$((\frac{1}{2!}+\frac{9}{10!})/2)*9$$, which clearly is not going to lend a correct answer. What was wrong with using the formula to find the sum of a sequence, in this case?

I guess you wanted to find the sum with the formula for evenly spaced sets: (first+last)/2*(# of terms). But the given sequence is not evenly spaced, thus you cannot apply it here.

Hope it's clear.

_________________

It takes time before all the things work together.

Math Expert
Joined: 02 Sep 2009
Posts: 36583
Followers: 7087

Kudos [?]: 93294 [0], given: 10555

Re: For all positive integers n, the sequence An is defined by [#permalink]

Show Tags

24 Jan 2014, 02:44
Intern
Status: Yes, I can and I will. 700+FTW
Joined: 30 Sep 2013
Posts: 20
Location: India
Concentration: Finance, Strategy
GMAT Date: 02-05-2014
GPA: 3.7
WE: Other (Retail)
Followers: 2

Kudos [?]: 15 [0], given: 3

Re: For all positive integers n, the sequence An is defined by [#permalink]

Show Tags

24 Jan 2014, 02:56
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13468
Followers: 575

Kudos [?]: 163 [0], given: 0

Re: For all positive integers n, the sequence An is defined by [#permalink]

Show Tags

06 Jun 2015, 06:50
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.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13468
Followers: 575

Kudos [?]: 163 [0], given: 0

Re: For all positive integers n, the sequence An is defined by [#permalink]

Show Tags

19 Jun 2016, 10:59
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.
_________________
Intern
Joined: 19 May 2016
Posts: 6
Followers: 0

Kudos [?]: 1 [0], given: 16

Re: For all positive integers n, the sequence An is defined by [#permalink]

Show Tags

08 Jul 2016, 07:17
An = (n-1)/n! = n/n! - 1/n! = 1/(n-1)! - 1/n!

A1 = 1/0! - 1/1!
A2 = 1/1! - 1/2!
A3 = 1/2! - 1/3!
.
.
A10 = 1/9! - 1/10!

As we see we can cancel out all terms except 1/0! - 1/10! = 1 - 1/10! = (10! - 1 )/10!
Re: For all positive integers n, the sequence An is defined by   [#permalink] 08 Jul 2016, 07:17
Similar topics Replies Last post
Similar
Topics:
2 For all positive integers n, the nth term in sequence S_n is defined 5 09 Sep 2015, 23:30
12 In the sequence gn defined for all positive integer values of n, g1 = 6 11 Jun 2015, 03:29
6 If sequence T is defined for all positive integers n such that tn +1 = 4 12 Jan 2015, 14:18
10 In the sequence g_n defined for all positive integer 6 22 Dec 2012, 06:28
29 The sequence f(n) = (2n)! ÷ n! is defined for all positive 4 15 Jan 2012, 14:28
Display posts from previous: Sort by