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Compute $\lfloor \sqrt{1} \rfloor + \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor + \dots + \lfloor \sqrt{100} \rfloor$.

 Apr 5, 2022
 #1
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+1

sumfor(n, 1, 100, floor(2#n))==625

 Apr 5, 2022
 #2
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\(\sqrt1+ ... \sqrt4+ ...\sqrt9+...\sqrt{16}+...\sqrt{25}+...\sqrt{36}+...\sqrt{49}+...\sqrt{64}+...\sqrt{81}+... \sqrt{100}\\ 3*1 + 5*2 + 7*3 + 9*4 + 11*5 + 13*6 + 15*7 + 17*8 + 19*9 + 10\\ 3+ 10 + 21 + 36 + 55 + 78 + 105 + etc \)

 Apr 5, 2022

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