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Method of Differences

  1. Nov 17, 2008 #1
    1. The problem statement, all variables and given/known data
    I'm given the identity:
    [tex]\sum{\frac{r^{3}-r+1}{r(r+1)} \equiv r-1 + \frac{1}{r} - \frac{1}{r+1}[/tex]

    EDIT: from r=1 to n on the sum sign
    2. Relevant equations

    I know I'm ment to use the method of differences and I know that

    [tex]\sum(r) = \frac{n(n+1)}{2}[/tex]


    3. The attempt at a solution

    I presume you need to find out how you can relate the 3 fractions to the summation of r?

    Thanks :)
     
    Last edited: Nov 17, 2008
  2. jcsd
  3. Nov 17, 2008 #2

    tiny-tim

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    Hi thomas49th! :smile:

    Hint:

    i] r3 - r = … ?

    ii] with a view to using differences, 1/r(r+1) = … - … ? :wink:
     
  4. Nov 17, 2008 #3
    sorry to of wasted your time but the orginal question was

    [tex]\sum{\frac{r^{3}-r+1}{r(r+1)} \equiv r-1 + \frac{1}{r} - \frac{1}{r+1}[/tex]

    there is no 1/2

    sorry!

    Errr can I combine it to this:
    [tex]\sum (r) - 1 + \sum(\frac{1}{r} - \frac{1}{r+1})[/tex]

    does that help at all?

    [tex]\sum(\frac{1}{r} - \frac{1}{r+1})[/tex]
    simplifies to [tex]\sum(\frac{1}{r^{2} + r}) [/tex]

    Thanks! :)
     
  5. Nov 17, 2008 #4

    tiny-tim

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    So ∑ 1/(r2 + r) = … ? :smile:
    erm … I could see it was wrong! … in fact, I still think it's wrong!! :rolleyes:
    Not following you. :confused:

    The method of difference involves subtracting the answer for r = n from the answer for r = n-1.
     
  6. Nov 17, 2008 #5
    It's question number 4 [​IMG]

    I could do part a) that's easy just part part b) is giving me trouble.

    At first I tried [tex]\sum (r) - 1 + \frac{1}{\sum (r)} - \frac{1}{\sum (r) + 1}[/tex]

    but that's wrong?

    Thanks :)
     
  7. Nov 17, 2008 #6

    tiny-tim

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    The ∑n part is right.

    The 1 should be ∑1.

    The rest is just ∑(1/r - 1/r+1) …

    that's already in differences, isn't it? …

    so its sum is … ? :smile:
     
  8. Nov 17, 2008 #7
    is the sum:
    [tex]\frac{n(n+1)}{2} - n + \frac{2}{n(n+1)} - \frac{2}{n(n+1) + 2n}[/tex]

    is that right?

    is the sum of 1 = n?
     
  9. Nov 17, 2008 #8

    tiny-tim

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    The first two terms are right. :smile:

    As I said before: The rest is just ∑(1/r - 1/r+1)

    Look … that's (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + …

    can you see what's happening?

    this is why it's called the method of differences, and why it works!
     
  10. Nov 17, 2008 #9
    I know the trick when things cancel out.
    So are you substituting values 1 to n into 1/2 - 1/r+1?

    how was I ment to spot that you substitute values into there and not the first part?

    Thnanks :)
     
  11. Nov 17, 2008 #10

    tiny-tim

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    Exactly! :biggrin:
    Because the first part was easy … it was just ∑(r + 1) …

    but the second part was ∑ 1/r(r+1), which obviously needs a trick! :smile:
     
  12. Nov 18, 2008 #11
    Ahh yes I got it

    [tex]\frac{n^3 + n}{2(n+1)}[/tex]

    Thank you ever so much :)
     
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