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Math Help - Summation

  1. #1
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    Summation

    Find \sum_{n=1}^n(\frac{1}{4r^2-8r+3}) given that
    \frac{2}{4r^2-8r+3}=\frac{1}{2r-3}-\frac{1}{2r-1}



    \sum_{n=1}^n(\frac{1}{4r^2-8r+3})=\frac{1}{2}\sum_{n= 1}^n(\frac{1}{2r-3}-\frac{1}{2r-1})

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

    =\frac{1}{2}[\frac{1}{\frac{n}{2}(-2+(n-1)(2))}]-\frac{1}{2}[\frac{1}{\frac{n}{2}(2-(n-1)(2))}]

    =\frac{1}{2}[\frac{1}{\frac{n}{2}(2n-4)}-\frac{1}{\frac{n}{2}(4-2n)}]

    = \frac{1}{2}[\frac{1}{\frac{n}{2}(2n-4)}+\frac{1}{\frac{n}{2}(2n-4)}]

    =\frac{1}{\frac{n}{2}(2n-4)}

    =\frac{1}{n(n-2)}

    but ans is -\frac{n}{2n-1}
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  2. #2
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    Quote Originally Posted by Punch View Post
    Find \sum_{n=1}^n(\frac{1}{4r^2-8r+3}) given that
    \frac{2}{4r^2-8r+3}=\frac{1}{2r-3}-\frac{1}{2r-1}



    \sum_{n=1}^n(\frac{1}{4r^2-8r+3})=\frac{1}{2}\sum_{n= 1}^n(\frac{1}{2r-3}-\frac{1}{2r-1})

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

    =\frac{1}{2}[\frac{1}{\frac{n}{2}(-2+(n-1)(2))}]-\frac{1}{2}[\frac{1}{\frac{n}{2}(2-(n-1)(2))}]

    =\frac{1}{2}[\frac{1}{\frac{n}{2}(2n-4)}-\frac{1}{\frac{n}{2}(4-2n)}]

    = \frac{1}{2}[\frac{1}{\frac{n}{2}(2n-4)}+\frac{1}{\frac{n}{2}(2n-4)}]

    =\frac{1}{\frac{n}{2}(2n-4)}

    =\frac{1}{n(n-2)}

    but ans is -\frac{n}{2n-1}
    You have taken, \sum_{r=1}^{n}\frac{1}{2r-3}=\frac{1}{\frac{n}{2}(-2+(n-1)(2))}.

    This is incorrect (substitute n=2 and the right hand side becomes undefined).

    Much easier is to treat the summation without breaking it apart into two pieces.

    You have to evaluate, \sum_{r=1}^n(\frac{1}{4r^2-8r+3})=\frac{1}{2}\sum_{r= 1}^n(\frac{1}{2r-3}-\frac{1}{2r-1})

    When r=1; \frac{1}{2r-3}-\frac{1}{2r-1}=\frac{1}{-1}-\frac{1}{1}

    When r=2; \frac{1}{2r-3}-\frac{1}{2r-1}=\frac{1}{1}-\frac{1}{3}

    When r=3; \frac{1}{2r-3}-\frac{1}{2r-1}=\frac{1}{3}-\frac{1}{5}
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    When r=n-1; \frac{1}{2r-3}-\frac{1}{2r-1}=\frac{1}{2n-5}-\frac{1}{2n-3}

    When r=n; \frac{1}{2r-3}-\frac{1}{2r-1}=\frac{1}{2n-3}-\frac{1}{2n-1}

    Adding all these together you will get the correct answer.
    Last edited by Sudharaka; June 9th 2011 at 09:39 PM.
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  3. #3
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    Quote Originally Posted by Sudharaka View Post
    You have taken, \sum_{r=1}^{n}\frac{1}{2r-3}=\frac{1}{\frac{n}{2}(-2+(n-1)(2))}.

    This is incorrect (substitute n=2 and the right hand side becomes undefined).

    Much easier is to treat the summation without breaking it apart into two pieces.

    You have to evaluate, \sum_{r=1}^n(\frac{1}{4r^2-8r+3})=\frac{1}{2}\sum_{r= 1}^n(\frac{1}{2r-3}-\frac{1}{2r-1})

    When r=1; \frac{1}{2r-3}-\frac{1}{2r-1}=\frac{1}{-1}-\frac{1}{1}

    When r=2; \frac{1}{2r-3}-\frac{1}{2r-1}=\frac{1}{1}-\frac{1}{3}

    When r=3; \frac{1}{2r-3}-\frac{1}{2r-1}=\frac{1}{3}-\frac{1}{5}
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    When r=n-1; \frac{1}{2r-3}-\frac{1}{2r-1}=\frac{1}{2n-5}-\frac{1}{2n-3}

    When r=n; \frac{1}{2r-3}-\frac{1}{2r-1}=\frac{1}{2n-3}-\frac{1}{2n-1}

    Adding all these together you will get the correct answer.
    \sum_{r=1}^n(\frac{1}{4r^2-8r+3})=\frac{1}{2}\sum_{r= 1}^n(\frac{1}{2r-3}-\frac{1}{2r-1})

    =-2+\frac{2}{3}+\frac{2}{15}+\frac{2}{35}

    but this is neither a GP nor AP
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  4. #4
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    wait, i think i see it, is this the method of differences?
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  5. #5
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    Quote Originally Posted by Punch View Post
    \frac{1}{2}\sum_{n=1}^n(\frac{1}{2r-3})-\frac{1}{2}\sum_{n= 1}^n(\frac{1}{2r-1})[/TEX]
    At this stage, you can just map the summands to each other, by, say, replacing r with r+1 for the first one:

     \displaystyle\begin{aligned} S & =   \frac{1}{2}\sum_{1 \le r \le n}\frac{1}{2r-3}-\frac{1}{2}\sum_{1 \le r \le n }^n\frac{1}{2r-1} \\& =   \frac{1}{2}\sum_{1 \le r+1 \le n}\frac{1}{2(r+1)-3}-\frac{1}{2}\sum_{1 \le r \le n }^n\frac{1}{2r-1} \\& =  \frac{1}{2}\sum_{0 \le r \le n-1}\frac{1}{2r-1}-\frac{1}{2}\sum_{1 \le r \le n }^n\frac{1}{2r-1} \\& = \frac{1}{2}\left(\frac{1}{2(0)-1}\right) -\frac{1}{2}\left(\frac{1}{2(n)-1}\right) + \frac{1}{2}\sum_{1 \le r \le n}\frac{1}{2r-1}-\frac{1}{2}\sum_{1 \le r \le n }^n\frac{1}{2r-1} \\& = \frac{1}{2}\bigg(\frac{1}{1-2n}-1\bigg).\end{aligned}

    Doing the above might seem bit hard/weird at first, but after few times it becomes so natural that you will never again want to write out terms and stuff! :]
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  6. #6
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    Quote Originally Posted by Punch View Post
    wait, i think i see it, is this the method of differences?
    Yes. Every term cancels with a term succeeding it. i.e: -1/1 cancels with 1/1, -1/3 cancels with 1/3 etc. Only the terms, -1 and -\frac{1}{2n-1} will be left.

    Hence \sum_{r=1}^{n}(\frac{1}{4r^2-8r+3})=\frac{1}{2}\left(-1-\frac{1}{2n-1}\right)
    Last edited by Sudharaka; June 10th 2011 at 01:51 AM. Reason: Forgot to put the minus sign.
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  7. #7
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    Hello, Punch!

    \text{Find }\,\sum_{n=1}^n\left(\frac{1}{4r^2-8r+3}\right)\,\text{ given that: }\:\frac{1}{4r^2-8r+3}\:=\:\frac{1}{2}\left(\frac{1}{2r-3}-\frac{1}{2r-1}\right)

    Write out the terms . . .

    S \;=\; \frac{1}{2}\sum^n_{r=1}\left(\frac{1}{2r-3} - \frac{1}{2r-1}\right)

    . . =\;\frac{1}{2}\bigg[\left(-\frac{1}{1} - \frac{1}{1}\right) + \left(\frac{1}{1} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \hdots + \left(\frac{1}{2n-3} - \frac{1}{2n-1}\right)\bigg]


    Inside the brackets, all terms cancel out except the first and last terms.

    . . S \;=\;\frac{1}{2}\left(-1 - \frac{1}{2n-1}\right) \;=\;\frac{1}{2}\left(\frac{-2n}{2n-1}\right)


    Therefore: . S \;=\;-\frac{n}{2n-1}

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