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

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Manager
Status: Never ever give up on yourself.Period.
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Kudos [?]: 77 [0], given: 35

For all positive integers n, the sequence An is defined by [#permalink]  22 Dec 2012, 05:34
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Difficulty:

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Question Stats:

46% (03:27) correct 54% (02:26) wrong based on 91 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 throughA_{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: 25207
Followers: 3419

Kudos [?]: 25105 [7] , given: 2702

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

_________________
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Kudos [?]: 34 [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 throughA_{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: 25207
Followers: 3419

Kudos [?]: 25105 [0], given: 2702

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 throughA_{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: 232
Concentration: Finance, Economics
GMAT 1: 670 Q39 V41
GMAT 2: 730 Q49 V41
Followers: 2

Kudos [?]: 34 [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 throughA_{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
Senior Manager
Joined: 03 Feb 2013
Posts: 275
Location: India
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GPA: 3.3
WE: Engineering (Computer Software)
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Kudos [?]: 7 [0], given: 279

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 throughA_{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
Never Give Up !!!

Intern
Status: Yes, I can and I will. 700+FTW
Joined: 30 Sep 2013
Posts: 20
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Concentration: Finance, Strategy
GMAT Date: 02-05-2014
GPA: 3.7
WE: Other (Retail)
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Kudos [?]: 7 [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 throughA_{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: 25207
Followers: 3419

Kudos [?]: 25105 [0], given: 2702

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 [?]: 7 [0], given: 3

Re: For all positive integers n, the sequence An is defined by [#permalink]  24 Jan 2014, 02:56
Re: For all positive integers n, the sequence An is defined by   [#permalink] 24 Jan 2014, 02:56
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