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Where Kn = * (1/n), and n is represented by a set of

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jimmyjamesdonkey
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Where Kn = * (1/n), and n is represented by a set of [#permalink] New post   Updated on: 12 May 2008, 13:07
Where Kn = [-1^(n-1)] * (1/n), and n is represented by a set of integers n = {1, 2, 3, 4, 5 . . . }, what must be true of the sum of the first 20 numbers in Sequence K?

Equal to -1
Equal to 1
Greater than 1
Less than 1
Less than 0



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Originally posted by jimmyjamesdonkey on 12 May 2008, 11:37.
Last edited by jimmyjamesdonkey on 12 May 2008, 13:07, edited 1 time in total.
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Re: Where Kn = * (1/n), and n is represented by a set of [#permalink] New post   12 May 2008, 12:34
jimmyjamesdonkey wrote:
Where Kn = -1(n-1) (1/n), and n is represented by a set of integers n = {1, 2, 3, 4, 5 . . . }, what must be true of the sum of the first 20 numbers in Sequence K?

Equal to -1
Equal to 1
Greater than 1
Less than 1
Less than 0


Kn = -1(n-1) (1/n) Are we multiplying these two??? b/c if we are its def less than 0 as every number will be negative.
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Re: Where Kn = * (1/n), and n is represented by a set of [#permalink] New post   12 May 2008, 13:08
Sorry...Edited my original post...should be correct now...
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Re: Where Kn = * (1/n), and n is represented by a set of [#permalink] New post   12 May 2008, 13:31
jimmyjamesdonkey wrote:
Where Kn = [-1^(n-1)] * (1/n), and n is represented by a set of integers n = {1, 2, 3, 4, 5 . . . }, what must be true of the sum of the first 20 numbers in Sequence K?

Equal to -1
Equal to 1
Greater than 1
Less than 1
Less than 0


looks to be less than 1..
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Re: Where Kn = * (1/n), and n is represented by a set of [#permalink] New post   12 May 2008, 13:56
Can you elaborate how you came to that conclusion?
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Re: Where Kn = * (1/n), and n is represented by a set of [#permalink] New post   12 May 2008, 14:04
If you start to calculate out the sequence, it's something like...

1, -1/2, 1/3, -1/4, etc.

You can group the sequence into groups of 2. The sum of each group is positive, but smaller than half of the previous group sum (with the first sum group = 1/2). Therefore, the total sum will be positive, but less than 1.
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Re: Where Kn = * (1/n), and n is represented by a set of [#permalink] New post   12 May 2008, 14:19
Would you normally try to group them with this type of problem, or is that just specific to this situation?
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Re: Where Kn = * (1/n), and n is represented by a set of [#permalink] New post   12 May 2008, 14:27
jimmyjamesdonkey wrote:
Where Kn = [-1^(n-1)] * (1/n), and n is represented by a set of integers n = {1, 2, 3, 4, 5 . . . }, what must be true of the sum of the first 20 numbers in Sequence K?

Equal to -1
Equal to 1
Greater than 1
Less than 1
Less than 0


-1^(n-1)*1/n Lets start off w/ the first number 1.

-1^(0)*1 --> 1

Next is -1/2
After that its 1/3
After that its -1/4


I would stop here, b/c going to 20 is obviously too time consuming. We can gather something here though 1-1/2+1/3-1/4=.58333

Here we can realize the numbers are going to get small and smaller. To the point where its almost insignificant. Id say D at this point.
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Re: Where Kn = * (1/n), and n is represented by a set of [#permalink] New post   12 May 2008, 14:28
I'm not sure. I just did it for this particular problem, because it helped me simplify and structure the problem.

Other cases I would group numbers would be adding up a series, let's say, and by pairing up numbers, you could find a pattern. An example would be adding up 1 thru 100. You can pair it up into 50 pairs of sum 101: (1+100), (2+99), (3+98), etc.
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Re: Where Kn = * (1/n), and n is represented by a set of [#permalink] New post   13 May 2008, 05:29
Just for kicks...Is there a way to sum of the 20 fractions quickly?



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Re: Where Kn = * (1/n), and n is represented by a set of [#permalink]
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