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

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

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New post 22 Dec 2012, 06:34
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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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New post 22 Dec 2012, 06:49
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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) \(\frac{10(10!)}{11!}\)
_____________________________________

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

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 \(\frac{10!-1}{10!}\).

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

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New post 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 \(\frac{10!-1}{10!}\).

Answer: C.



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

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New post 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 \(\frac{10!-1}{10!}\).

Answer: C.



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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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

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New post 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 \(\frac{10!-1}{10!}\).

Answer: C.



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

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New post 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 \(\frac{10!-1}{10!}\).

Answer: C.



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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New post 24 Jan 2014, 03:44
rgyanani wrote:
are there more such type of questions bunnel?


Yes:
n-is-an-integer-greater-than-or-equal-to-0-the-sequence-tn-110969.html
sequence-a-is-defined-by-the-equation-an-3n-2-where-n-114912.html
the-next-number-in-a-certain-sequence-is-defined-by-multiply-159626.html
the-sequence-of-numbers-a1-a2-a3-an-is-defined-by-an-154049.html
the-sequence-f-n-2n-n-is-defined-for-all-positive-126126.html
sequence-s-is-defined-as-follows-s1-2-s2-2-1-s3-2-2-sn-129435.html
the-infinite-sequence-s-k-is-defined-as-s-k-10-s-k-1-k-126033.html
the-infinite-sequence-a1-a2-an-is-defined-such-that-an-104906.html
the-nth-term-tn-of-a-certain-sequence-is-defined-as-tn-t-105320.html

Theory on sequences problems: sequences-progressions-101891.html

All DS sequences problems to practice: search.php?search_id=tag&tag_id=111
All PS sequences problems to practice: search.php?search_id=tag&tag_id=112


You might also find the following post helpful: new-project-review-discuss-and-get-kudos-points-153555.html

And finally, you can check our questions banks for all questions by category: viewforumtags.php
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Re: For all positive integers n, the sequence An is defined by  [#permalink]

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New post 24 Jan 2014, 03:56
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Re: For all positive integers n, the sequence An is defined by  [#permalink]

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New post 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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New post 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 \(\frac{10!-1}{10!}\).

Answer: C.


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

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New post 27 Apr 2017, 02:02
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 \(\frac{10!-1}{10!}\).

Answer: C.


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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Re: For all positive integers n, the sequence An is defined by &nbs [#permalink] 24 Jul 2018, 06:23
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