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Math Help - Trival question.

  1. #1
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    Trival question.

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

    I realized that


    This is the same thing as

    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 ...


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

     k < n

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

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

    Is there a clearer way to show that there are exactly (n/2) number of 2n's .
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by ╔(σ_σ)╝ View Post
    1 +3 +5 + ... (2n-1) = n^{2}
    How do we show this.

    I realized that


    This is the same thing as

    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 ...


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

     k < n

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

    Leaving me with \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:

    \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...

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

    So 2S=2n\cdot n \implies S=n^2
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  3. #3
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     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.
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  4. #4
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    Thanks guys.
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