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Method of differences exam question [ ]

  1. Jan 29, 2009 #1
    Method of differences exam question [urgent]

    1. The problem statement, all variables and given/known data

    Now i know that this is equal to [tex]\frac{2}{r} + \frac{1}{r+1} -\frac{3}{r+2}[/tex]

    I need to use the method of differences to work out the summation of the series from 1 to n.

    so i substitute values in


    [tex]2 + \frac{1}{2} - \frac{3}{3}[/tex]


    [tex]1 + \frac{1}{3} - \frac{3}{4}[/tex]


    [tex]\frac{2}{3} + \frac{1}{4} - \frac{3}{5}[/tex]


    [tex]\frac{2}{n-1} + \frac{1}{n} - \frac{3}{n+1}[/tex]


    [tex]\frac{2}{n} + \frac{1}{n+1} - \frac{3}{n+2}[/tex]

    But I cant see any terms that have cancelled out. How do i simplify this

    Thanks :)
  2. jcsd
  3. Jan 29, 2009 #2


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    Re: Method of differences exam question [urgent]

    r=n+1, (-3/(n+3))+1/(n+2)+2/(n+1). Look at the 2/(n+1) from this, the 1/(n+1) from r=n and the -3/(n+1) from r=n-1.
  4. Jan 29, 2009 #3


    Staff: Mentor

    Re: Method of differences exam question [urgent]

    To clarify, by "this" in your second line, you meant the summand, not the series sum.

    You have [tex]\sum_{r = 1}^n (\frac{2}{r} + \frac{1}{r + 1} - \frac{3}{r + 2})[/tex]
    Break this sum into three separate summations. After that, move constant multiples outside the summation. Now, align the terms of all three summations by pulling out enough terms so that all of the sums start with the same fraction. I think you'll see that you get a lot of cancellation then.
  5. Jan 29, 2009 #4
    Re: Method of differences exam question [urgent]

    Yup got it. Thanks alot! diagonal 3!

    Cheers :)
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