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Math Help - Summation: Method of differences.

  1. #1
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    Summation: Method of differences.

    Heres the question:

    Show that
    \frac{2}{r-1} - \frac{1}{r+ 1} - \frac{1}{r + 2} = \frac{3r + 4}{ r(r+1)(r+2)}

    I did this part since its simple algebra. .

    Then the second part i'm stuck.

    ii.) Hence write an expression in terms of n for
     \sum \frac{3r + 4}{ r(r+1)(r+2)} to n. Where it starts r = 1

    Okey, so, method of differences.

    I start by putting in r = 1

    \frac{2}{1} - \frac{1}{ 2} - \frac{1}{3}

    Then r = 2

    \frac{2}{ 2} - \frac{1}{3} - \frac{1}{4}

    Then r = n-1

    \frac{2}{n-1} - \frac{1}{n} - \frac{1}{n+1}

    Then r = n

    \frac{2}{n} - \frac{1}{n+1} - \frac{1}{n+2}

    Then r = r + 1

    \frac{2}{n+1} - \frac{1}{n+2} - \frac{1}{n+3}

    So then you add them all. Some cancel and im left with

    \frac{2}{1} + \frac{2}{2} + \frac{2}{n-1} + \frac{2}{n} + \frac{2}{n+1}

    Then i simplify and i get

    2 + 1 + \frac{2n(3n - 1)}{ n(n+1)(n-1)}

    And this is incorrect for the answer is

    2 + 1 - \frac{1}{2} - \frac{2}{n+1} - \frac{1}{n+2}

    Thankyou for any help you can offer.

    How do i do all these maths things to make it easier to read?

    EDIT: Nvm i found the tutorial . I'll edit this with it now.
    Last edited by AshleyT; May 20th 2008 at 04:12 AM.
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  2. #2
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    Quote Originally Posted by AshleyT View Post
    Heres the question:

    Show that
    \frac{2}{r-1} - \frac{1}{r+ 1} - \frac{1}{r + 2} = \frac{3r + 4}{ r(r+1)(r+2)}

    I did this part since its simple algebra. .
    How did you do this? It is wrong!

    The correct identity is

    \frac{2}{r} - \frac{1}{r+ 1} - \frac{1}{r + 2} = \frac{3r + 4}{ r(r+1)(r+2)}

    Then the second part i'm stuck.

    ii.) Hence write an expression in terms of n for
     \sum \frac{3r + 4}{ r(r+1)(r+2)} to n. Where it starts r = 1

    Okey, so, method of differences.\
     \sum_{r=1}^{r=n} \frac{3r + 4}{ r(r+1)(r+2)}

    = \sum_{r=1}^{r=n} \frac{2}{r} - \frac{1}{r+ 1} - \frac{1}{r + 2}

    = \sum_{r=1}^{r=n} 2\left(\frac{1}{r} - \frac{1}{r+ 1}\right) +\left(\frac{1}{r+1} - \frac{1}{r + 2}\right)


    = \sum_{r=1}^{r=n} 2\left(\frac{1}{r} - \frac{1}{r+ 1}\right) + \sum_{r=1}^{r=n} \left(\frac{1}{r+1} - \frac{1}{r + 2}\right)


    = \sum_{r=1}^{r=n} 2\left(\frac{1}{r} - \frac{1}{r+ 1}\right) + \sum_{r=2}^{r= n+1} \left(\frac{1}{r} - \frac{1}{r + 1}\right)


    = 2 - \frac2{n+1} + \frac12 - \frac1{n+2}

    = 2 + 1 - \frac{1}{2} - \frac{2}{n+1} - \frac{1}{n+2}<br />

    EDIT: Oh yay , I got the right answer
    And so on.... I might have made arithmetic mistakes... but I hope you get the general idea.
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  3. #3
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    Quote Originally Posted by Isomorphism View Post
    How did you do this? It is wrong!

    The correct identity is

    \frac{2}{r} - \frac{1}{r+ 1} - \frac{1}{r + 2} = \frac{3r + 4}{ r(r+1)(r+2)}



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

    = \sum_{r=1}^{r=n} \frac{2}{r} - \frac{1}{r+ 1} - \frac{1}{r + 2}

    = \sum_{r=1}^{r=n} 2\left(\frac{1}{r} - \frac{1}{r+ 1}\right) +\left(\frac{1}{r+1} - \frac{1}{r + 2}\right)


    = \sum_{r=1}^{r=n} 2\left(\frac{1}{r} - \frac{1}{r+ 1}\right) + \sum_{r=1}^{r=n} \left(\frac{1}{r+1} - \frac{1}{r + 2}\right)


    = \sum_{r=1}^{r=n} 2\left(\frac{1}{r} - \frac{1}{r+ 1}\right) + \sum_{r=2}^{r= n+1} \left(\frac{1}{r} - \frac{1}{r + 1}\right)


    = 2 - \frac2{n+1} + \frac12 - \frac1{n+2}

    = 2 + 1 - \frac{1}{2} - \frac{2}{n+1} - \frac{1}{n+2}<br />

    EDIT: Oh yay , I got the right answer
    And so on.... I might have made arithmetic mistakes... but I hope you get the general idea.
    Oops, i meant 2/r sorry. Typed it out wrong!

    How did you go from



     \sum_{r=1}^{r=n} \frac{2}{r} - \frac{1}{r+ 1} - \frac{1}{r + 2}<br />

    to

    <br />
 \sum_{r=1}^{r=n} 2\left(\frac{1}{r} - \frac{1}{r+ 1}\right) +\left(\frac{1}{r+1} - \frac{1}{r + 2}\right)<br />

    because, i can see the 2\frac{1}{r} but not how - \frac{1}{r+ 1} goes in there?


    Tanks
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  4. #4
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    Quote Originally Posted by AshleyT View Post
    Oops, i meant 2/r sorry. Typed it out wrong!

    How did you go from



    " alt="= \sum_{r=1}^{r=n} \frac{2}{r} - \frac{1}{r+ 1} - \frac{1}{r + 2}
    " />

    to


    " alt="= \sum_{r=1}^{r=n} 2\left(\frac{1}{r} - \frac{1}{r+ 1}\right) +\left(\frac{1}{r+1} - \frac{1}{r + 2}\right)
    " />

    because, i can see the
    2\left(\frac{1}{r} but not how - \frac{1}{r+ 1}\right) goes in there?

    Tanks


    I just wrote -\frac{1}{r+1} as -\frac2{r+1} + \frac1{r+1} and then grouped -\frac2{r+1} with \frac2{r} and \frac1{r+1} with \frac1{r+2}
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  5. #5
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    Quote Originally Posted by Isomorphism View Post
    I just wrote -\frac{1}{r+1} as -\frac2{r+1} + \frac1{r+1} and then grouped -\frac2{r+1} with \frac2{r} and \frac1{r+1} with \frac1{r+2}

    O i see.

    Thankyou very much
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