May 13th, 2019, 01:07 AM  #11 
Newbie Joined: Sep 2017 From: Belgium Posts: 19 Thanks: 7 
And if you want to find the sum of the integer parts instead of the integer part of the sum, there is also this: $\sum\limits_{n\leq x}\lfloor n^{\frac{1}{k}} \rfloor=\lfloor x^{\frac{1}{k}} \rfloor (x+1)\sum\limits_{n\leq x^{\frac{1}{k}}} n^k$ 
May 13th, 2019, 07:33 AM  #12 
Global Moderator Joined: Dec 2006 Posts: 20,926 Thanks: 2205 
For 6 terms, one finds $\displaystyle \lfloor y \rfloor = 10 < 1 + \!\int_1^6\! \sqrt{x} dx \approx 10.131$.

May 13th, 2019, 12:35 PM  #13 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra 
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). https://www.desmos.com/calculator/jfd5ybw1fz 

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