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Homework Help: Sum of the square roots of the first n natural numbers

  1. Aug 2, 2006 #1
    Is there a way to find the,"Sum of the square roots of the first n natural numbers"?
  2. jcsd
  3. Aug 2, 2006 #2


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    I don't think you could do it exactly. You could approximate it by the integral of [itex]\sqrt{x}[/itex], and get a bound on the error.
  4. Aug 2, 2006 #3
    As StatusX says I'm pretty sure there's no way to do in exactly in closed-form. If you don't have a way to calculate square roots at all (ie. you're doing it without a calculator and don't want to go through an approximation method), then a simple integer approximation would be

    [tex]\frac{2}{3}\lfloor \sqrt{n} \rfloor^3 - \frac{1}{2}\lfloor \sqrt{n} \rfloor^2 - \frac{1}{6} \lfloor \sqrt{n} \rfloor + \lfloor \sqrt{n} \rfloor(n-\lfloor \sqrt{n} \rfloor^2),[/tex]

    but it's not very good. The integral approximation [itex]\frac{2}{3} n^{\frac{3}{2}}[/itex] is much better, but you have to be able to compute [itex]n^{3/2}[/itex]
    ([itex]2/3 \lfloor n^{3/2}\rfloor[/itex] is also better than the one I gave above though).
    Last edited: Aug 2, 2006
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