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Thread: 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:

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

    It's implied that I should use that:

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

    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
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    Proof by induction would work here.

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

    Base case: True.

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

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

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

    Simplify $\displaystyle {\color{red}(1)}$ and notice that in $\displaystyle {\color{red}(2)}$:
    $\displaystyle \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}.}$$\displaystyle \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; Aug 13th 2008 at 02:02 PM.
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  3. #3
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    Ahh...that's a way.



    Which yields

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

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

    Thanks.
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  4. #4
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    Hello, Aslak!

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

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

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

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


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

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


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

    . . $\displaystyle 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] $

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


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

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


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

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