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Subject: "Interesting sequence"     Previous Topic | Next Topic
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solo
Member since Jul-31-04
Jul-31-04, 10:10 AM (EST)
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"Interesting sequence"
 
   Could someone please show me how to find the answer to the following sum (without actually using a calculator):

1/(2x3) + 1/(3x4) + 1/(4x5) + ... + 1/(99x11)

Thanks.


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alexb
Charter Member
1359 posts
Jul-31-04, 10:12 AM (EST)
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1. "RE: Interesting sequence"
In response to message #0
 
   >Could someone please show me how to find the answer to the
>following sum (without actually using a calculator):
>
>1/(2x3) + 1/(3x4) + 1/(4x5) + ... + 1/(99x11)
>

This is a part of what is known as the telescoping series. Use the identity

1/(n×(n-1)) = 1/(n-1) - 1/n.


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solo
Member since Jul-31-04
Jul-31-04, 11:45 AM (EST)
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2. "RE: Interesting sequence"
In response to message #1
 
   Thanks for the answer. It's so simple if you know how!

Joe


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Bruce Campbell
guest
Oct-27-04, 12:49 PM (EST)
 
3. "RE: Interesting sequence"
In response to message #1
 
   This is really stupid and bugging the heck out of me. Here I am a 44-yr-old man and not a total moron, but I'm not seeing how to get from 1/(nx(n-1)) to 1/(n-1) - 1/n.

Can you tell me if the actual steps are published anywhere that demonstrate how to get from point a to point b? I can even do it backwards, no problem, but don't see the intuitive method of getting from one to the other in the example indicated above.

Came across your site in a google search for the telescoping problem listed above in an effort to help my 7th grader. Yeah, big help, eh?


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alexb
Charter Member
1359 posts
Oct-27-04, 01:10 PM (EST)
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4. "RE: Interesting sequence"
In response to message #3
 
   >I'm not
>seeing how to get from 1/(nx(n-1)) to 1/(n-1) - 1/n.

>
>Can you tell me if the actual steps are published anywhere
>that demonstrate how to get from point a to point b? I can
>even do it backwards, no problem, but don't see the
>intuitive method of getting from one to the other in the
>example indicated above.

Getting from b to a is as good as getting from a to b. The two are equal, after all.

However, there is a method of getting from a to b directly. I am sure you would rather prefer going from b to a. The pattern there is quite simple:

1/6 = 1/2 - 1/3
1/12 = 1/3 - 1/4, etc.

However, here's the method. It's called the method of "undetermined coefficients." It starts with an assertion that a fraction, like k/(m×n) can always be split into a sum of two fractions with denominators m and n. The numerators are to be found. Assume

k/(m×n) = x/m + y/n.

Then

k/(m×n) = (xn + ym)/(m×n),

so that

k = xn + ym;

You are of course looking for integer solutions. The only numbers k that can be represented in such a form are multiples of gcd(m,n). (You may want to search this site in particular or anywhere on the web for Euclid's algorithm.) In the case of the telescoping sequence, m = n-1, and, obviously, gcd(n-1,n) = 1. Thus you know that there are x and y such that

1 = xn + y(n-1).

But what might they be? It's almost immediate that x=1 and y=-1. In a more formal way you can try to use an "extension of Euclid's algorithm" (search this site.)


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Sarah
guest
Oct-29-04, 10:27 AM (EST)
 
5. "RE: Interesting sequence"
In response to message #1
 
   >This is a part of what is known as the telescoping series. Use the >identity
>
>1/(n×(n-1)) = 1/(n-1) - 1/n.

That is an interesting series but I am curious how the general term
1/(nx(n-1))
gives the last term in the sum
1/(99x11).


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