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May 13th, 2019, 01:07 AM   #11
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And if you want to find the sum of the integer parts instead of the integer part of the sum, there is also this:
$\sum\limits_{n\leq x}\lfloor n^{\frac{1}{k}} \rfloor=\lfloor x^{\frac{1}{k}} \rfloor (x+1)-\sum\limits_{n\leq x^{\frac{1}{k}}} n^k$
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May 13th, 2019, 07:33 AM   #12
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For 6 terms, one finds $\displaystyle \lfloor y \rfloor = 10 < 1 + \!\int_1^6\! \sqrt{x} dx \approx 10.131$.
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May 13th, 2019, 12:35 PM   #13
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I can't see how you are excluding 30. This is what I imagine your working to be like (although I'm a bit lost as to where the 2+ comes from).
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