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Math Help - The value of x

  1. #1
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    The value of x



    what is the value of x?
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  2. #2
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    Re: The value of x

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  3. #3
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    Re: The value of x

    Quote Originally Posted by Mhmh96 View Post


    what is the value of x?
    x=\frac{1}{\sqrt 2} \cdot \displaystyle \sum_{n=0}^{1003} \frac{1}{\sqrt n +\sqrt{n+1}}
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    Re: The value of x

    Helpful answers,thanks !
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  5. #5
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    Re: The value of x

    FOR k = 0 to 2006 STEP 2
    x = x + 1 / [SQRT(k) + SQRT(k+2)]
    NEXT k

    x = 22.4053565...

    SURPRISE(?): x^2 = 502 (integer)
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  6. #6
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    Re: The value of x

    It appears that $\sum\frac{1}{\sqrt{2n}+\sqrt{2(n+1)}}$from n = 0 to k is \sqrt{(k+1)/2}
    It is almost certain that x = $\sqrt{502}$
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  7. #7
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    Re: The value of x

    Well, for all the good this'll do:

    1 / [SQRT(n) + SQRT(n + 2)] = SQRT{[n + 1 - SQRT(n^2 + 2n)] / 2}
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  8. #8
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    Re: The value of x

    I'm surprised no one was able to figure this out until now (including me). It's actually fairly simple.
    x=\frac{1}{\sqrt 2} \cdot \displaystyle \sum_{n=0}^{1003} \frac{1}{\sqrt n +\sqrt{n+1}}
    x=\frac{1}{\sqrt 2} \cdot \displaystyle \sum_{n=0}^{1003} \frac{\sqrt n -\sqrt{n+1}}{(\sqrt n -\sqrt{n+1})(\sqrt n +\sqrt{n+1})}
    x=\frac{1}{\sqrt 2} \cdot \displaystyle \sum_{n=0}^{1003} \sqrt{n+1} -\sqrt n
    Now it is a telescoping series.
    x=\frac{1}{\sqrt 2}(\sqrt{1003+1} -\sqrt 0)
    x= \sqrt{502}
    Last edited by browni3141; April 9th 2012 at 06:51 PM. Reason: Added Tex tags
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