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705-805 (Hard)|   Arithmetic|      
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daviesj
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For all positive integers n, the sequence An is defined by 22 Dec 2012, 06:34
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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55% (02:46) correct 45% (03:05) wrong based on 902 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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C
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For all positive integers n, the sequence An is defined by 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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For all positive integers n, the sequence An is defined by 08 Jul 2016, 08:17
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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!
General Discussion
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For all positive integers n, the sequence An is defined by 24 Nov 2013, 13:55
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Bunuel
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.
Show more


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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For all positive integers n, the sequence An is defined by 24 Nov 2013, 14:04
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AccipiterQ
Bunuel
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?
Show more

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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For all positive integers n, the sequence An is defined by 24 Nov 2013, 14:23
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Bunuel
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Bunuel
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.
Show more


oh my goodness, I completely overlooked that in my zeal to be clever
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For all positive integers n, the sequence An is defined by 20 Jan 2014, 09:26
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daviesj
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) \(\frac{10(10!)}{11!}\)
Show more

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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For all positive integers n, the sequence An is defined by 24 Jan 2014, 03:31
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are there more such type of questions bunnel?

Bunuel
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Bunuel
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.
Show more
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Bunuel
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For all positive integers n, the sequence An is defined by 24 Jan 2014, 03:44
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rgyanani
are there more such type of questions bunnel?
Show more

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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For all positive integers n, the sequence An is defined by 26 Apr 2017, 22:22
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Bunuel
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.
Show more

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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For all positive integers n, the sequence An is defined by 27 Apr 2017, 02:02
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Bunuel
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?
Show more

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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For all positive integers n, the sequence An is defined by 17 May 2019, 06:19
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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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For all positive integers n, the sequence An is defined by 17 May 2019, 06:25
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09173140521
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 !?
Show more
because it is factorial !!! :cool:
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SPatel1992
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For all positive integers n, the sequence An is defined by 17 May 2019, 15:41
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kinjiGC
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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) \(\frac{10(10!)}{11!}\)


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) \(\frac{10(10!)}{11!}\)
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- 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!
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shubhim20
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For all positive integers n, the sequence An is defined by 11 Apr 2025, 02:37
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isn't n/2 (2a + (n-1)d )applicable here?

Bunuel
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Bunuel
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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shubhim20
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For all positive integers n, the sequence An is defined by 11 Apr 2025, 02:40
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still can't get how did you get 3+2
Bunuel
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Bunuel
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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For all positive integers n, the sequence An is defined by 11 Apr 2025, 03:05
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shubhim20
still can't get how did you get 3+2

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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Do it another way:

1/2! + 2/3! = 1/2 + 2/6 = 1/2 + 1/3 = (3 + 2)/6 = 5/6.
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For all positive integers n, the sequence An is defined by 12 Apr 2026, 03:06
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