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Summation of series using method of difference

  1. Aug 17, 2012 #1
    1. The problem statement, all variables and given/known data
    Here's my question. My school recently taught me finding summation using method of difference and what my teacher taught was just involving 2 partial fractions.

    But this question appeared in my exercise given by my teacher.

    r th term: (2r-1)/r(r+1)(r+2). Find summation of n th terms begin with r=1.

    Can someone show me how to solve this using method of difference?


    2. The attempt at a solution
    What I got is sum (- 1/2r + 3/r+1 - 5/2(r+2) ).

    And I stucked. What I used to know is r th term is f(r) - f(r-1). and sum of rth term is f(n)- f(0).
    I cant get the f(r) - f(r-1) either because -1/2r and -5/2(r+2) both having the same sign.
     
  2. jcsd
  3. Aug 17, 2012 #2

    tiny-tim

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    welcome to pf!

    hi hhm28! welcome to pf! :smile:
    might be easier to split it up first

    for example 2/(r+1)(r+2) - 1/r(r+1)(r+2) :wink:
     
  4. Aug 17, 2012 #3
    This did help me. But the problem is i stucked somewhere.

    I got sigma -[(2/r+2) - (3/r+1) + (1/2r+2) + (1/2r)] and I stucked.

    I should've make it into f(r)-f(r-1). =( I'm stupid Argh....

    Can you show me how? Perhaps upload a photo. LOL
     
  5. Aug 17, 2012 #4

    tiny-tim

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    hint: how would you do ∑ 1/r(r+1) ?

    use the same method for ∑ 1/r(r+1)(r+2) :smile:
     
  6. Aug 17, 2012 #5
    LOL. Thats the problem. I only know how to solve f(r) - f(r-1).

    Now there is 3 partial fractions. But with 2/(r+1)(r+2) - 1/r(r+1)(r+2), i managed to get ∑ -[(2/r+2) - (3/r+1) + (1/2r+2) + (1/2r)].
     
  7. Aug 17, 2012 #6

    tiny-tim

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    ok, so how would you do ∑ 1/r(r+1) ? :smile:
     
  8. Aug 17, 2012 #7
    1/r(r+1) = (1/r) - (1/r+1)

    ∑ 1/r(r+1) = ∑ (1/r) - (1/r+1)
    = -∑ [(1/r+1) -(1/r)]

    Let f(r)= 1/ r+1 , f(r-1)= 1/r
    ∑ 1/r(r+1) = -∑ f(r) - f(r-1)
    = - [f(n) - f(0)]
    = - [(1/n+1) - 1]
    = n/n+1
     
  9. Aug 17, 2012 #8

    tiny-tim

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    ok, now do the same for ∑ 1/r(r+1)(r+2) :smile:
     
  10. Aug 17, 2012 #9
    ∑1/r(r+1)(r+2) = ∑1/2r - 1/(r+1) + 1/2(r+2)

    How do i continue? the 1/(r+1) is disrupting. =(
     
  11. Aug 17, 2012 #10

    tiny-tim

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    no, try double fractions like 1/r(r+1) :wink:
     
  12. Aug 17, 2012 #11
    means that A/r(r+1) + B/(r+2) to get partial?
     
  13. Aug 17, 2012 #12

    tiny-tim

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    A/r(r+1) + B/(r+1)(r+2) :smile:
     
  14. Aug 17, 2012 #13
    Gosh. I never learned that =(
     
  15. Aug 17, 2012 #14
    Thank you so much anywhere. =)
     
  16. Aug 19, 2012 #15

    vela

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    Good start. Try writing out the first few terms:

    r=1: ##-\frac{1}{2}\left(\frac{1}{1}\right) + 3\left(\frac{1}{2}\right) - \frac{5}{2}\left(\frac{1}{3}\right)##

    r=2: ##-\frac{1}{2}\left(\frac{1}{2}\right) + 3\left(\frac{1}{3}\right) - \frac{5}{2}\left(\frac{1}{4}\right)##

    r=3: ##-\frac{1}{2}\left(\frac{1}{3}\right) + 3\left(\frac{1}{4}\right) - \frac{5}{2}\left(\frac{1}{5}\right)##

    r=4: ##-\frac{1}{2}\left(\frac{1}{4}\right) + 3\left(\frac{1}{5}\right) - \frac{5}{2}\left(\frac{1}{6}\right)##

    r=5: ##-\frac{1}{2}\left(\frac{1}{5}\right) + 3\left(\frac{1}{6}\right) - \frac{5}{2}\left(\frac{1}{7}\right)##

    Now look at the pieces with (1/3) (or (1/4) or (1/5)). Can you see a pattern to how the various parts cancel?
     
    Last edited: Aug 19, 2012
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