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Math Help - Method of Difference

  1. #1
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    Method of Difference

    Find \sum_{r = 1}^n(\frac{1}{r}+\frac{3}{r+1}-\frac{4}{r+2})

    \sum_{r = 1}^n(\frac{1}{r}+\frac{3}{r+1}-\frac{4}{r+2})=\sum_{r = 1}^n\frac{1}{r}+3\sum_{r = 1}^n\frac{1}{r+1}-4\sum_{r = 1}^n\frac{1}{r+2}
    Last edited by Punch; June 11th 2011 at 12:03 AM.
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  2. #2
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    Sum ranges from r =1, not n =1, right? It's equivalent to:

     \sum_{r=-1}^{n-2}\frac{1}{r+2}+3\sum_{r=0}^{n-1}\frac{1}{r+2}-4\sum_{r=1}^{n}\frac{1}{r+2}
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  3. #3
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    Quote Originally Posted by TheCoffeeMachine View Post
    Sum ranges from r =1, not n =1, right? It's equivalent to:

     \sum_{r=-1}^{n-2}\frac{1}{r+2}+3\sum_{r=0}^{n-1}\frac{1}{r+2}-4\sum_{r=1}^{n}\frac{1}{r+2}
    sorry, mistake on my part
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  4. #4
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    \begin{aligned} S & = \sum_{1 \le r \le n}\frac{1}{r}+3\sum_{1 \le r \le n}\frac{1}{r+1}-4\sum_{1 \le r \le n}\frac{1}{r+2} \\& = \sum_{1 \le r+2 \le n}\frac{1}{r+2}+3\sum_{1 \le r +1\le n}\frac{1}{r+2}-4\sum_{1 \le r \le n}\frac{1}{r+2} \\& = \sum_{-1 \le r \le n-2}\frac{1}{r+2}+3\sum_{0 \le r\le n-1}\frac{1}{r+2}-4\sum_{1 \le r \le n}\frac{1}{r+2} \\& = \frac{1}{(-1)+2}+\frac{1}{(0)+2}-\frac{1}{(n-1)+2}-\frac{1}{(n)+2} +\frac{3}{(0)+2}-\frac{3}{(n)+2} \\& +\sum_{1 \le r \le n}\frac{1}{r+2}+3\sum_{1 \le r \le n}\frac{1}{r+2}-4\sum_{1 \le r \le n}\frac{1}{r+2} \\& = 3-\frac{1}{n+1}-\frac{4}{n+2}+4\sum_{1 \le r \le n}\frac{1}{r+2}-4\sum_{1 \le r \le n}\frac{1}{r+2} \\& = \boxed{3-\frac{1}{n+1}-\frac{4}{n+2}}.\end{aligned}
    Last edited by TheCoffeeMachine; June 11th 2011 at 01:09 AM.
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  5. #5
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    Hello, Punch!

    \text{Find: }\;S \;=\;\sum_{r = 1}^n \left(\frac{1}{r}+\frac{3}{r+1}-\frac{4}{r+2}\right)

    Write out the terms . . .

    S \;=\;\left(\frac{1}{1} + \frac{3}{2} - \frac{4}{3}\right) + \left(\frac{1}{2} + \frac{3}{3} - \frac{4}{4}\right) + \left(\frac{1}{3} + \frac{3}{4} - \frac{4}{5}\right) + \left(\frac{1}{4} + \frac{3}{5} - \frac{4}{6}\right) +\; \hdots

    . . . .  \hdots \;+ \left(\frac{1}{n-2} + \frac{3}{n-1} - \frac{4}{n}\right) + \left(\frac{1}{n-1} + \frac{3}{n} - \frac{4}{n+1}\right) + \left(\frac{1}{n} + \frac{3}{n+1} - \frac{4}{n+2}\right)


    We note that most of the terms cancel out . . .

    S \;=\;\left(\frac{1}{1} + \frac{3}{2} - \rlap{/}\frac{4}{3}\right) + \left(\frac{1}{2} + \rlap{/}\frac{3}{3} - \rlap{/}\frac{4}{4}\right) + \left(\rlap{/}\frac{1}{3} + \rlap{/}\frac{3}{4} - \rlap{/}\frac{4}{5}\right) + \left(\rlap{/}\frac{1}{4} + \rlap{/}\frac{3}{5} - \rlap{/}\frac{4}{6}\right) +\; \hdots

    . . . .  \hdots\; + \left(\rlap{/////}\frac{1}{n-2} + \rlap{/////}\frac{3}{n-1} - \rlap{/}\frac{4}{n}\right) + \left(\rlap{/////}\frac{1}{n-1} + \rlap{/}\frac{3}{n} - \frac{4}{n+1}\right) + \left(\rlap{/}\frac{1}{n} + \frac{3}{n+1} - \frac{4}{n+2}\right)


    And we have: . S \;=\;\frac{1}{1} + \frac{3}{2} + \frac{1}{2} - \frac{4}{n+1} + \frac{3}{n+1} - \frac{4}{n+2}

    . . . . . . . . . . . S \;=\;3 - \frac{1}{n+1} - \frac{4}{n+2} \;=\;\frac{3n^2+4n}{(n+1)(n+2)}


    Therefore:. S \;=\;\frac{n(3n+4)}{(n+1)(n+2)}

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