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Method of differences exam question

  1. Nov 30, 2008 #1
    1. The problem statement, all variables and given/known data

    given the identity [tex]\frac{2}{(r+1)(r+3)} \equiv \frac{1}{r+1} - \frac{1}{r+3}[/tex]

    prove that

    [tex] \sum^{n}_{r = 1} \frac{2}{(r+1)(r+3)} = \frac{n(an + b)}{6(n+2)(n+3)}[/tex]

    where a and b are constants to be found

    2. Relevant equations

    3. The attempt at a solution

    I subed in number r = 1,2,3,n

    r =1: 1/2 - 1/4
    r=2:1/3 - 1/5
    r= 3: 1/4-1/6
    r = n: 1/(n+1) - 1/(n+3)

    the ones in red being the ones that dont cancel... so i add them but i get 5n-9/6(n+3)

    is there somthing else i need to do?
    Last edited: Nov 30, 2008
  2. jcsd
  3. Nov 30, 2008 #2


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    Homework Helper

    you are making this hard
    given the partial fractions hint
    write out the inverses of the sums posive one row ngative the next
    so past the first few terms we net 2 positive and 2 negative terms

    in short you forgot -1/(n+2) does not cancel
    Last edited: Nov 30, 2008
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