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

 It is currently 13 Oct 2015, 14:36

### 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:
Manager
Status: Never ever give up on yourself.Period.
Joined: 23 Aug 2012
Posts: 147
Location: India
Concentration: Finance, Human Resources
Schools: MBS '17 (A)
GMAT 1: 570 Q47 V21
GMAT 2: 690 Q50 V33
GPA: 3.5
WE: Information Technology (Investment Banking)
Followers: 9

Kudos [?]: 184 [1] , given: 35

For all positive integers n, the sequence An is defined by [#permalink]  22 Dec 2012, 05:34
1
KUDOS
3
This post was
BOOKMARKED
00:00

Difficulty:

85% (hard)

Question Stats:

48% (03:19) correct 52% (02:22) wrong based on 130 sessions
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) : beat-it-no-one-wants-to-be-defeated-journey-570-to-149968.html

Math Expert
Joined: 02 Sep 2009
Posts: 29856
Followers: 4924

Kudos [?]: 53965 [7] , given: 8260

Re: For all positive integers n, the sequence An is defined by [#permalink]  22 Dec 2012, 05:49
7
KUDOS
Expert's post
1
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 $$\frac{10!-1}{10!}$$.

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

Kudos [?]: 76 [0], given: 40

Re: For all positive integers n, the sequence An is defined by [#permalink]  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: 29856
Followers: 4924

Kudos [?]: 53965 [0], given: 8260

Re: For all positive integers n, the sequence An is defined by [#permalink]  24 Nov 2013, 13:04
Expert's post
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.
_________________
Current Student
Joined: 26 Sep 2013
Posts: 221
Concentration: Finance, Economics
GMAT 1: 670 Q39 V41
GMAT 2: 730 Q49 V41
Followers: 2

Kudos [?]: 76 [0], given: 40

Re: For all positive integers n, the sequence An is defined by [#permalink]  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
Director
Joined: 03 Feb 2013
Posts: 845
Location: India
Concentration: Operations, Strategy
Schools: Tepper '17 (M)
GMAT 1: 760 Q49 V44
GPA: 3.88
WE: Engineering (Computer Software)
Followers: 73

Kudos [?]: 494 [0], given: 537

Re: For all positive integers n, the sequence An is defined by [#permalink]  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
Struggling with GMAT ? Experience http://www.gmatify.com/

My Application Experience : http://gmatclub.com/forum/hardwork-never-gets-unrewarded-for-ever-189267-40.html#p1516961

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 [?]: 10 [0], given: 3

Re: For all positive integers n, the sequence An is defined by [#permalink]  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: 29856
Followers: 4924

Kudos [?]: 53965 [0], given: 8260

Re: For all positive integers n, the sequence An is defined by [#permalink]  24 Jan 2014, 02:44
Expert's post
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 [?]: 10 [0], given: 3

Re: For all positive integers n, the sequence An is defined by [#permalink]  24 Jan 2014, 02:56
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 6784
Followers: 369

Kudos [?]: 83 [0], given: 0

Re: For all positive integers n, the sequence An is defined by [#permalink]  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.
_________________
Re: For all positive integers n, the sequence An is defined by   [#permalink] 06 Jun 2015, 06:50
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
6 In the sequence gn defined for all positive integer values of n, g1 = 5 11 Jun 2015, 03:29
5 If sequence T is defined for all positive integers n such that tn +1 = 3 12 Jan 2015, 14:18
5 In the sequence g_n defined for all positive integer 3 22 Dec 2012, 06:28
11 The sequence f(n) = (2n)! ÷ n! is defined for all positive 2 15 Jan 2012, 14:28
Display posts from previous: Sort by