Results 1 to 9 of 9

Thread: Series

  1. #1
    Member SengNee's Avatar
    Joined
    Jan 2008
    From
    Pangkor Island, Perak, Malaysia.
    Posts
    155

    Series

    Show that, when $\displaystyle 2n$ terms of the series

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

    are grouped in pairs, an arithmetic series is formed.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    Quote Originally Posted by SengNee View Post
    Show that, when $\displaystyle 2n$ terms of the series

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

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

    $\displaystyle u_r=(2r-1)^2-(2r)^2=(2r)^2-4r+1-(2r)^2=1-4r$

    so:

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

    RonL
    Last edited by CaptainBlack; Jan 20th 2008 at 11:20 PM. Reason: Correct (2k)^2 typo
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    Quote Originally Posted by SengNee View Post
    Show that, when $\displaystyle 2n$ terms of the series

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

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

    $\displaystyle u_r=(2r-1)^2-(2r)^2=(2r)^2-4r+1-(2r)^2=1-4r$

    so:

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

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

    $\displaystyle u_{r+1} - u_r = .....$ therefore ....
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member SengNee's Avatar
    Joined
    Jan 2008
    From
    Pangkor Island, Perak, Malaysia.
    Posts
    155
    Quote Originally Posted by CaptainBlack View Post
    Grouping the terms gives pairs of the form:

    $\displaystyle u_r=(2r-1)^2-(2r)^2=(2r)^2-4r+1-(2r)^2=1-4r$

    so:

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

    RonL
    $\displaystyle \sum_{k=1}^{2n}(-1)^{k+1}(2k)^2$
    I don't understand this part...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    Quote Originally Posted by SengNee View Post
    $\displaystyle \sum_{k=1}^{2n}(-1)^{k+1}(2k)^2$
    I don't understand this part...
    It's a shorthand way of writing $\displaystyle 1^2-2^2+3^2-4^2+...+(2r-1)^2-(2r)^2+...+(2n-1)^2-(2n)^2$.

    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, SengNee!

    Here's a rather primitive approach . . .


    Show that, when $\displaystyle 2n$ terms of the series

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

    are grouped in pairs, an arithmetic series is formed.

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

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


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


    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member SengNee's Avatar
    Joined
    Jan 2008
    From
    Pangkor Island, Perak, Malaysia.
    Posts
    155
    Quote Originally Posted by mr fantastic View Post
    It's a shorthand way of writing $\displaystyle 1^2-2^2+3^2-4^2+...+(2r-1)^2-(2r)^2+...+(2n-1)^2-(2n)^2$.

    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.
    $\displaystyle \sum_{k=1}^{2n}(-1)^{k+1}(2k)^2$
    $\displaystyle \sum_{k=1}^{2n}(-1)^{k+1}(k)^2$

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

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

    For the second one, when $\displaystyle k=2n$,

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


    Isn't it?
    Last edited by SengNee; Jan 20th 2008 at 10:18 PM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member SengNee's Avatar
    Joined
    Jan 2008
    From
    Pangkor Island, Perak, Malaysia.
    Posts
    155
    Quote Originally Posted by Soroban View Post
    Hello, SengNee!

    Here's a rather primitive approach . . .



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

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


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


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


    I do like this. Is it correct also?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member SengNee's Avatar
    Joined
    Jan 2008
    From
    Pangkor Island, Perak, Malaysia.
    Posts
    155
    $\displaystyle Can \quad anyone \quad help \quad me({\color{blue}\text{8th}} \quad post)?$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: Oct 3rd 2011, 01:12 AM
  2. Replies: 3
    Last Post: Sep 29th 2010, 06:11 AM
  3. Replies: 0
    Last Post: Jan 26th 2010, 08:06 AM
  4. Replies: 2
    Last Post: Dec 1st 2009, 12:45 PM
  5. Replies: 1
    Last Post: May 5th 2008, 09:44 PM

Search Tags


/mathhelpforum @mathhelpforum