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Math Help - Series

  1. #1
    Member SengNee's Avatar
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    Series

    Show that, when 2n terms of the series

    1^2-2^2+3^2-4^2+...+(2r-1)^2-(2r)^2+...+(2n-1)^2-(2n)^2

    are grouped in pairs, an arithmetic series is formed.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by SengNee View Post
    Show that, when 2n terms of the series

    1^2-2^2+3^2-4^2+...+(2r-1)^2-(2r)^2+...+(2n-1)^2-(2n)^2

    are grouped in pairs, an arithmetic series is formed.
    Grouping the terms gives pairs of the form:

    u_r=(2r-1)^2-(2r)^2=(2r)^2-4r+1-(2r)^2=1-4r

    so:

    \sum_{k=1}^{2n}(-1)^{k+1}k^2 = \sum_{r=1}^n (1-4r)

    RonL
    Last edited by CaptainBlack; January 20th 2008 at 11:20 PM. Reason: Correct (2k)^2 typo
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  3. #3
    Flow Master
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    Quote Originally Posted by SengNee View Post
    Show that, when 2n terms of the series

    1^2-2^2+3^2-4^2+...+(2r-1)^2-(2r)^2+...+(2n-1)^2-(2n)^2

    are grouped in pairs, an arithmetic series is formed.
    Quote Originally Posted by CaptainBlack View Post
    Grouping the terms gives pairs of the form:

    u_r=(2r-1)^2-(2r)^2=(2r)^2-4r+1-(2r)^2=1-4r

    so:

    \sum_{k=1}^{2n}(-1)^{k+1}(2k)^2 = \sum_{r=1}^n (1-4r)

    RonL
    So to explicitly address the show bit, now that the hard work's been done:

    u_{r+1} - u_r = ..... therefore ....
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  4. #4
    Member SengNee's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Grouping the terms gives pairs of the form:

    u_r=(2r-1)^2-(2r)^2=(2r)^2-4r+1-(2r)^2=1-4r

    so:

    \sum_{k=1}^{2n}(-1)^{k+1}(2k)^2= \sum_{r=1}^n (1-4r)

    RonL
    \sum_{k=1}^{2n}(-1)^{k+1}(2k)^2
    I don't understand this part...
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  5. #5
    Flow Master
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    Quote Originally Posted by SengNee View Post
    \sum_{k=1}^{2n}(-1)^{k+1}(2k)^2
    I don't understand this part...
    It's a shorthand way of writing 1^2-2^2+3^2-4^2+...+(2r-1)^2-(2r)^2+...+(2n-1)^2-(2n)^2.

    If you haven't met this notation before, don't worry about it.

    If you want to research it a bit, it's called summation or sigma notation.
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  6. #6
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    Hello, SengNee!

    Here's a rather primitive approach . . .


    Show that, when 2n terms of the series

    1^2-2^2+3^2-4^2+...+(2r-1)^2-(2r)^2+ \cdots +(2n-1)^2-(2n)^2

    are grouped in pairs, an arithmetic series is formed.

    Group in pairs: . (1^2-2^2) + (3^2-4^2) + (5^2-6^2) + (7^2-8^2) + \cdots + \left([2n-1]^2 - [2n]^2\right)

    . . . . . . . . . = \quad(-3)\quad+\quad(-7) \;\;\;+\;\;\; (-11) \;\;+\;\; (-15)\quad + \cdots +\quad \left(-[4n-1]\right)


    This is an arithmetic series with first term a = \text{-}3 and common difference d = \text{-}4


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  7. #7
    Member SengNee's Avatar
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    Quote Originally Posted by mr fantastic View Post
    It's a shorthand way of writing 1^2-2^2+3^2-4^2+...+(2r-1)^2-(2r)^2+...+(2n-1)^2-(2n)^2.

    If you haven't met this notation before, don't worry about it.

    If you want to research it a bit, it's called summation or sigma notation.
    \sum_{k=1}^{2n}(-1)^{k+1}(2k)^2
    \sum_{k=1}^{2n}(-1)^{k+1}(k)^2

    I think the first one is false, the second one is correct.
    Because for the first one, when k=2n,

    u_{2n}
    =(-1)^{2n+1}(4n)^2
    =-(4n)^2

    For the second one, when k=2n,

    u_{2n}
    =(-1)^{2n+1}(2n)^2
    =-(2n)^2


    Isn't it?
    Last edited by SengNee; January 20th 2008 at 10:18 PM.
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  8. #8
    Member SengNee's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, SengNee!

    Here's a rather primitive approach . . .



    Group in pairs: . (1^2-2^2) + (3^2-4^2) + (5^2-6^2) + (7^2-8^2) + \cdots + \left([2n-1]^2 - [2n]^2\right)

    . . . . . . . . . = \quad(-3)\quad+\quad(-7) \;\;\;+\;\;\; (-11) \;\;+\;\; (-15)\quad + \cdots +\quad \left(-[4n-1]\right)


    This is an arithmetic series with first term a = \text{-}3 and common difference d = \text{-}4


    S_{2n}=1^2-2^2+3^2-4^2+...+(2n-1)^2-(2n)^2
    S_{2n}=-(2n)^2+(2n-1)^2-(2n-2)^2+(2n-3)^2+...-2^2+1^2
    2S_{2n}=(1-4n^2)+(4n^2-4n-3)+(5+8n-4n^2)+ (4n^2-12n-7)+...+(4n^2-4n-3)+(1-4n^2)
    2S_{2n}=(-4n-2)+(-4n-2)+...+(-4n-2)
    2S_{2n}=2n(-4n-2)
    S_{2n}=n(-4n-2)
    S_{2n}=-2n(2n+1)


    I do like this. Is it correct also?
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  9. #9
    Member SengNee's Avatar
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    Can \quad anyone \quad help \quad me({\color{blue}\text{8th}} \quad post)?
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