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Math Help - Acceptable proof?

  1. #1
    Member Jones's Avatar
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    Acceptable proof?

    Hi,

    Is this an acceptable proof of induction?

    show that, for all n >= 2

    \frac{1}{2}n^{3/2} \textless \sum_{k=1}^n\sqrt{k}~\textless~n^{3/2}

    multiply by 2 and verify the statement holds for n=2

    \sqrt{2^3} = 2\sqrt2 ~\textless~ 2\sum_{k=1}^n 1 + \sqrt{2} ~\textless~ 2\sqrt{8}


    The inductive step:

    n^{3/2}+(n+1)^{1/2} \rightarrow n^2 +1 \textless \sum_{k=1}^{n+1}\sqrt{k}~\textless~2n^{3/2} + (n+1)^{1/2} \rightarrow 2n^2 +1

    Acceptable?
    Last edited by Jones; October 15th 2009 at 11:42 AM.
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  2. #2
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    It would be a lot easier (for me, at least) to help you if you gave more detail on your steps...

    I have no idea how you ended up with that for n=2 ( \sqrt{2^3} \neq \sqrt{2}...) or what you did in the final step...

    for n=2 it should be:

    \frac{\sqrt{2^3}}{2}=\sqrt{\frac{2^3}{2^2}} = \sqrt{2}

    \sum_{k=1}^{2}\sqrt{k} = 1 + \sqrt{2}> \sqrt{2}

    \sqrt{2^3}=\sqrt{8} = 2\sqrt{2} > 1 + \sqrt{2}

    So \sqrt{2}<1+\sqrt{2}<2\sqrt{2} is true...

    Now please explain in more detail the final step
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  3. #3
    Member Jones's Avatar
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    Quote Originally Posted by Defunkt View Post
    It would be a lot easier (for me, at least) to help you if you gave more detail on your steps...

    I have no idea how you ended up with that for n=2 ( \sqrt{2^3} \neq \sqrt{2}...) or what you did in the final step...

    for n=2 it should be:

    \frac{\sqrt{2^3}}{2}=\sqrt{\frac{2^3}{2^2}} = \sqrt{2}

    \sum_{k=1}^{2}\sqrt{k} = 1 + \sqrt{2}> \sqrt{2}

    \sqrt{2^3}=\sqrt{8} = 2\sqrt{2} > 1 + \sqrt{2}

    So \sqrt{2}<1+\sqrt{2}<2\sqrt{2} is true...

    Now please explain in more detail the final step
    Hi,
    Sorry it was a typo
    I think you figured it out, was i meant was that  2^{3/2} = \sqrt{2^3} = 2\sqrt{2}
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