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Thread: Summation

  1. #1
    Member helloying's Avatar
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    Summation

    Show that $\displaystyle \sum (-1)^{r+1} r^2 $ = -n(2n+1)


    summation is from r=1 to 2n . sry i dont know how to type

    Thx
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  2. #2
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    Quote Originally Posted by helloying View Post
    Show that $\displaystyle \sum (-1)^{r+1} r^2 $ = -n(2n+1)


    summation is from r=1 to 2n . sry i dont know how to type

    Thx
    hi

    \sum^{2n}_{r=1}(-1)^{r+1} r^2 will generate

    $\displaystyle \sum^{2n}_{r=1}(-1)^{r+1} r^2$

    $\displaystyle =1-4+9-16+25-36+...$

    $\displaystyle =1^2-2^2+3^2-4^2+...+(2n-1)^2-(2n)^2$

    $\displaystyle =(1+2)(1-2)+(3+4)(3-4)+...+(2n+2n-1)(2n-2n-1)$

    $\displaystyle =-3-7-...-(4n-1)$

    which is an AP with first term -3 , last term -(4n-1)

    $\displaystyle S_n=\frac{n}{2}(-3-(4n-1))$

    simplify this and you are done.
    Last edited by mathaddict; May 27th 2010 at 11:37 PM. Reason: algebraic error
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  3. #3
    Senior Member oldguynewstudent's Avatar
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    Quote Originally Posted by mathaddict View Post
    hi

    \sum^{2n}_{r=1}(-1)^{r+1} r^2 will generate

    $\displaystyle \sum^{2n}_{r=1}(-1)^{r+1} r^2$

    $\displaystyle =1-4+9-16+25-36+...$

    $\displaystyle =1^2-2^2+3^2-4^2+...+(2n)^2-(2n-1)^2$

    $\displaystyle =(1+2)(1-2)+(3+4)(3-4)+...+(2n+2n-1)(2n-2n+1)$

    $\displaystyle =-3-4-...-(4n-1)$

    which is an AP with first term -3 , last term -(4n-1)

    $\displaystyle S_n=\frac{n}{2}(-3-(4n-1))$

    simplify this and you are done.
    I didn't quite follow this.

    How did you get $\displaystyle =1^2-2^2+3^2-4^2+...+(2n)^2-(2n-1)^2$?

    When I do this I come up with $\displaystyle =1^2-2^2+3^2-4^2+...-(2n)^2 +(2n-1)^2$, because $\displaystyle (-1)^{2n+1} {2n}^2$ = $\displaystyle -{2n}^2$.
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  4. #4
    Member helloying's Avatar
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    Quote Originally Posted by oldguynewstudent View Post
    I didn't quite follow this.

    How did you get $\displaystyle =1^2-2^2+3^2-4^2+...+(2n)^2-(2n-1)^2$?

    When I do this I come up with $\displaystyle =1^2-2^2+3^2-4^2+...-(2n)^2 +(2n-1)^2$, because $\displaystyle (-1)^{2n+1} {2n}^2$ = $\displaystyle -{2n}^2$.
    yep i realise i dont get it. i agree with oldguynewstudnet.and also in the part $\displaystyle
    =(1+2)(1-2)+(3+4)(3-4)+...+(2n+2n-1)(2n-2n+1)
    $


    it shd be -3-7 and not -3-4
    Last edited by Plato; May 24th 2010 at 04:47 AM.
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  5. #5
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    Quote Originally Posted by helloying View Post
    yep i realise i dont get it. i agree with oldguynewstudnet.and also in the part $\displaystyle
    =(1+2)(1-2)+(3+4)(3-4)+...+(2n+2n-1)(2n-2n+1)
    $

    it shd be -3-7 and not -3-4
    yup , some algebraic error here and there . Edited .
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