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

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22 Dec 2012, 06:34
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Question Stats:

54% (02:48) correct 46% (03:01) wrong based on 269 sessions

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

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

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22 Dec 2012, 06:49
12
6
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 $$\frac{10!-1}{10!}$$.

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

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24 Nov 2013, 13:55
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: 58335
Re: For all positive integers n, the sequence An is defined by  [#permalink]

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24 Nov 2013, 14:04
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.
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Re: For all positive integers n, the sequence An is defined by  [#permalink]

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24 Nov 2013, 14:23
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
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Re: For all positive integers n, the sequence An is defined by  [#permalink]

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20 Jan 2014, 09:26
1
1
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..
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Thanks,
Kinjal

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

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

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24 Jan 2014, 03:31
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.

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Math Expert
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Posts: 58335
Re: For all positive integers n, the sequence An is defined by  [#permalink]

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24 Jan 2014, 03:44
Intern
Status: Yes, I can and I will. 700+FTW
Joined: 30 Sep 2013
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Re: For all positive integers n, the sequence An is defined by  [#permalink]

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24 Jan 2014, 03:56
Intern
Joined: 19 May 2016
Posts: 6
Re: For all positive integers n, the sequence An is defined by  [#permalink]

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08 Jul 2016, 08:17
1
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!
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Re: For all positive integers n, the sequence An is defined by  [#permalink]

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26 Apr 2017, 22:22
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'm having a hard time keeping up with how you get this: $$A_1+A_2=\frac{1}{2!}=\frac{2!-1}{2!}$$ and this: $$A_1+A_2+A_3=\frac{5}{3!}=\frac{3!-1}{3!}$$

Can you break it down step by step?
- Specifically, for A1+A2, why do you bring the factorial to the numerator? And what makes it 2!-1?
- For A1+A2+A3, how do you get "5" for the numerator?
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Joined: 02 Sep 2009
Posts: 58335
Re: For all positive integers n, the sequence An is defined by  [#permalink]

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27 Apr 2017, 02:02
1
LakerFan24 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'm having a hard time keeping up with how you get this: $$A_1+A_2=\frac{1}{2!}=\frac{2!-1}{2!}$$ and this: $$A_1+A_2+A_3=\frac{5}{3!}=\frac{3!-1}{3!}$$

Can you break it down step by step?
- Specifically, for A1+A2, why do you bring the factorial to the numerator? And what makes it 2!-1?
- For A1+A2+A3, how do you get "5" for the numerator?

1. $$A_1+A_2=\frac{1}{2!}$$ but if you notice, you could write 1 as 2! - 1.

2. $$A_1+A_2+A_3=0+\frac{1}{2!}+\frac{2}{3!}=\frac{3+2}{3!}=\frac{5}{3!}=\frac{3!-1}{3!}$$.

Hope it's clear.
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Re: For all positive integers n, the sequence An is defined by  [#permalink]

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17 May 2019, 06:19
anybody can explain why we cannot use this formula to solve this problem ?
sum of sequence term= n/2 * (last+first) ? i do it but got different answer !?
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Re: For all positive integers n, the sequence An is defined by  [#permalink]

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17 May 2019, 06:25
09173140521 wrote:
anybody can explain why we cannot use this formula to solve this problem ?
sum of sequence term= n/2 * (last+first) ? i do it but got different answer !?

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

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17 May 2019, 15:41
kinjiGC wrote:
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) $$\frac{10(10!)}{11!}$$
- 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..[/quote]

This is a really good solution!
Re: For all positive integers n, the sequence An is defined by   [#permalink] 17 May 2019, 15:41
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