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Math Help - Formula for sum of squared odd numbers.

  1. #1
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    Formula for sum of squared odd numbers.

    Hello!

    I'm having trouble proving this relation:

    1^2+3^2+5^2+...(2n+1)^2=\frac{(n+1)(2n+1)(2n+3)}{3  }

    It's implied that I should use that:

    1^2+2^2+3^2+...+k^2=\frac{k(k+1)(2k+1)}{6}

    I've tried substituting k=(2n+1) in the second formula, but it's not working. I can't really see why it isn't...

    Could someone lead me in the right direction here?
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  2. #2
    o_O
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    Proof by induction would work here.

    Let P_{n} be the statement that: 1^2+3^2+5^2+...(2n+1)^2=\frac{(n+1)(2n+1)(2n+3)}{3  }

    Base case: True.

    Inductive step: Assuming P_{k} is true, it remains to show that P_{k+1} is also true, that is: 1^2+3^2+5^2+...(2(k+1)+1)^2=\frac{((k+1)+1)(2(k+1)  +1)(2(k+1)+3)}{3} \qquad {\color{red}(1)}

    Modifying it a bit just to see what we're working with:
    \text{LHS} = 1^2 + 3^2 + ... + (2k + 3)^2
    \text{LHS} = \underbrace{1^2 + 3^2 + ... + (2k + 1)^2}_{P_{k}} + (2k+3)^2 (I just expanded the '...' a bit and noticed that P_{k} is 'in' there)
    \text{LHS} = \frac{(k+1)(2k+1)(2k+3)}{3} + (2k+3)^2 \qquad{\color{red}(2)}

    Now we have to show that the right hand side of {\color{red}(1)} and {\color{red}(2)} are the same.

    Simplify {\color{red}(1)} and notice that in {\color{red}(2)}:
    \frac{(k+1)(2k+1)(2k+3)}{3} + (2k+3)^2 \: \: = \: \: \frac{(k+1)(2k+1)(2k+3) + 3(2k+3)^2}{3} \: \: = \: \: {\color{white}.} \frac{(2k+3) \bigg[(k+1)(2k+1) + 3(2k+3)\bigg]}{3}

    And now it is a matter of simplifying the inside the brackets.
    Last edited by o_O; August 13th 2008 at 02:02 PM.
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  3. #3
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    Ahh...that's a way.



    Which yields

    \frac{(2k+3)(k+2)(2k+5)}{3}

    Which is equivalent to the P_n-statement for the case n=k+1.

    Thanks.
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  4. #4
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    Hello, Aslak!

    Prove: . S \;=\;1^2+3^2+5^2+\hdots +(2n+1)^2 \;=\;\frac{(n+1)(2n+1)(2n+3)}{3}

    It's implied that I should use: . 1^2+2^2+3^2+\hdots+k^2\:=\:\frac{k(k+1)(2k+1)}{6}

    S is the sum of the squares of the odd integers up to (2n+1)^2

    Hence, it is the sum of the squares of all the integers up to (2n+1)^2
    . . minus the sum of the squares of the even integers up to (2n)^2.


    The sum of all squares up to (2n+1)^2 is:

    . . S_{\text{all}} \;=\;\frac{(2n+1)(2n+2)(4n +3)}{6} \;=\;\frac{(n+1)(2n+1)(4n+3)}{3}


    The sum of even squares up to (2n)^2 is:

    . . S_{\text{even}} \;=\;2^2 + 4^2 + 6^2 + \hdots + (2n)^2 \;= \;2^2\left[1^2 + 2^2 + 3^2 + \hdots + n^2\right]

    . . . . . . = \;4\cdot\frac{n(n+1)(2n+1)}{6} \;=\;\frac{2n(n+1)(2n+1)}{3}


    Then: . S \;=\;S_{\text{all}} - S_{\text{even}} \;=\;\frac{(n+1)(2n+1)(4n+3)}{3} - \frac{2n(n+1)(2n+1)}{3}

    . . Factor: . S \;=\;\frac{(n+1)(2n+1)}{3}\bigg[(4n+3) - 2n\bigg]


    Therefore: . \boxed{S \;=\;\frac{(n+1)(2n+1)(2n+3)}{3}}

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  5. #5
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    Now I fully understand. Thank you.
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