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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

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?

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.

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

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.. _________________

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.

_________________

It takes time before all the things work together.

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. _________________

Low GPA MBA Acceptance Rate Analysis Many applicants worry about applying to business school if they have a low GPA. I analyzed the low GPA MBA acceptance rate at...

To hop from speaker to speaker, to debate, to drink, to dinner, to a show in one night would not be possible in most places, according to MBA blogger...