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Thread: Quickie #4

  1. #1
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    Quickie #4


    \text{Given: }\;\frac{1^3 + 3^3 + 5^3 + \cdots + (2n-1)^3}{2^3 + 4^3 + 6^3 + \cdots + (2n)^3} \;=\;\frac{199}{242}

    . . \text{find }n.

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  2. #2
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    \sum_{k=1}^{n}(2k-1)^{3}=2n^{4}-n^{2}

    \sum_{k=1}^{n}(2k)^{3}=2n^{2}(n+1)^{2}

    \frac{2n^{4}-n^{2}}{2n^{2}(n+1)^{2}}=\frac{2n^{2}-1}{2(n+1)^{2}}

    \frac{2n^{2}-1}{2(n+1)^{2}}=\frac{199}{242}

    .
    .
    .
    .

    86n^{2}-796n-640=0

    2(n-10)(43n+32)=0

    \boxed{n=10} \;\ or \;\ \frac{-32}{43}
    Last edited by galactus; Dec 28th 2006 at 08:50 AM.
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  3. #3
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    Nice solution, Galactus!

    By the way, these "Quickie" solutions are not meant to be "better" than yours.
    Most of them employ a "trick" of some sort ... some well-known, others obscure.
    I found each to be an eye-opening experience.



    \text{Given: }\;\frac{1^3 + 3^3 + 5^3 + \cdots + (2n-1)^3}{2^3 + 4^3 + 6^3 + \cdots + (2n)^3} \;=\;\frac{199}{242}

    . . \text{find }n.

    We are expected to know: . \sum^n_{k=1}k^3\:=\:\left[\sum^n_{k=1}k\right]^2\:=\:\frac{n^2(n+1)^2}{4}

    . . and that: .If \frac{a}{b} = \frac{c}{d}, then: . \frac{a+b}{b} \:=\:\frac{c+d}{d}


    So we have: . \frac{1^3 + 2^3 + 3^3 + \cdots + (2n)^3}{2^3(1^3 + 2^3 + 3^3 + \cdots + n^3)} \;=\;\frac{199+242}{242}

    Then: . \frac{\frac{(2n)^2(2n+1)^2}{4} }{2^3\!\cdot\!\frac{n^2(n+1)^2}{4}} \:=\:\frac{441}{242} \quad\Rightarrow\quad \frac{4n^2(2n+1)^2}{8n^2(n+1)^2} \:=\:\frac{21^2}{2(11^2)}

    . . Hence: . \left(\frac{2n+1}{n+1}\right)^2 \:=\:\left(\frac{21}{11}\right)^2


    It concludes with: "Clearly, n = 10.
    . . The negative square root gives a negative value for n."

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