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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]

Show Tags

20 Jan 2014, 09:26

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]

Show Tags

06 Jun 2015, 07:50

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).

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