Results 1 to 4 of 4

Thread: Trival question.

  1. #1
    Banned
    Joined
    Feb 2009
    From
    Posts
    79

    Trival question.

    $\displaystyle 1 +3 +5 + ... (2n-1) = n^{2}$
    How do we show this.

    I realized that


    This is the same thing as

    $\displaystyle 1 +3 +5 + ... + (2n-5) + (2n-3)+(2n-1) = n^{2}$

    And the 1's , 3's and 5's cancel out but i'm left with a bunch of 2n's

    Here is my reasoning ...


    $\displaystyle 1 +3 +5 + ......+ k + (2n-k) ....... + (2n-5) + (2n-3)+(2n-1) = n^{2}$

    $\displaystyle k < n$

    And i assumed that there must be $\displaystyle \frac{n}{2} 2n's $

    Leaving me with $\displaystyle \frac{n}{2} 2n = n^{2} $

    Is there a clearer way to show that there are exactly (n/2) number of 2n's .
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Quote Originally Posted by ╔(σ_σ)╝ View Post
    $\displaystyle 1 +3 +5 + ... (2n-1) = n^{2}$
    How do we show this.

    I realized that


    This is the same thing as

    $\displaystyle 1 +3 +5 + ... + (2n-5) + (2n-3)+(2n-1) = n^{2}$

    And the 1's , 3's and 5's cancel out but i'm left with a bunch of 2n's

    Here is my reasoning ...


    $\displaystyle 1 +3 +5 + ......+ k + (2n-k) ....... + (2n-5) + (2n-3)+(2n-1) = n^{2}$

    $\displaystyle k < n$

    And i assumed that there must be $\displaystyle \frac{n}{2} 2n's $

    Leaving me with $\displaystyle \frac{n}{2} 2n = n^{2} $

    Is there a clearer way to show that there are exactly (n/2) number of 2n's .
    In summation notation, this is just:

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

    If you want to do it your way, you can do what Gauss did...

    $\displaystyle 1~~~~~~~+3~~~~~~~+...+2n-3+2n-1$
    $\displaystyle \underline{2n-1+2n-3+...+3~~~~~~~+1~~~~~~~~~~+}$

    So $\displaystyle 2S=2n\cdot n \implies S=n^2$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Jan 2009
    Posts
    715
    $\displaystyle 2k + 1 = (k+1)^2 - k^2 $

    Then solve it by method of differences
    Last edited by simplependulum; May 9th 2009 at 10:00 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Feb 2009
    From
    Posts
    79
    Thanks guys.
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum