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Question Number 190067 by DAVONG last updated on 26/Mar/23

Answered by mehdee42 last updated on 26/Mar/23

$$S=\left[\sqrt{1}\right]+\left[\sqrt{2}\right]+...+\left[\sqrt{500}\right]$$$$\left[\sqrt{1}\right]+\left[\sqrt{2}\right]+\left[\sqrt{3}\right]=3\times 1$$$$\left[\sqrt{4}\right]+\left[\sqrt{5}\right]+\left[6\right]+\left[\sqrt{7}\right]+\left[\sqrt{8}\right]=5\times 2$$$$⋮$$$$\left[\sqrt{441}\right]+\left[\sqrt{442}\right]+...+\left[\sqrt{483}\right]=43\times 21$$$$\left[\sqrt{484}\right]+\left[\sqrt{485}\right]+...+\left[\sqrt{500}\right]=27\times 22$$$$\Rightarrow S=\underset{1}{\sum ^{21}}(2i+1)i+27\times 22=2\underset{1}{\sum ^{21}}{i}^2+\underset{1}{\sum ^{21}}i+374$$$$=2\times \frac{21\times 22\times 43}{6}+\frac{21\times 22}{2}+374=7227$$

Commented by DAVONG last updated on 27/Mar/23

$$\mathrm{Thanks}\mathrm{sire}!$$

Answered by mr W last updated on 26/Mar/23

$$S_n=\left[\sqrt{1}\right]+\left[\sqrt{2}\right]+\left[\sqrt{3}\right]+...+\left[\sqrt{n}\right]$$$$letm=\lfloor \sqrt{n}\rfloor$$$$\left[\sqrt{1}\right]+\left[\sqrt{2}\right]+\left[\sqrt{3}\right]=3\times 1$$$$\left[\sqrt{4}\right]+\left[\sqrt{5}\right]+\left[\sqrt{6}\right]+\left[\sqrt{7}\right]+\left[\sqrt{8}\right]=5\times 2$$$$\left[\sqrt{9}\right]+\left[\sqrt{10}\right]+...+\left[\sqrt{15}\right]=7\times 3$$$$...$$$$\left[\sqrt{k^2}\right]+\left[\sqrt{k^2+1}\right]+...+\left[\sqrt{k^2+2k}\right]=(2k+1)\times k$$$$...$$$$\left[\sqrt{m^2}\right]+\left[\sqrt{m^2+1}\right]+...+\sqrt{n}=(n+1-m^2)\times m$$$$$$$$S_n=\underset{k=1}{\sum ^{m-1}}(2k+1)k+(n+1-m^2)m$$$$S_n=2\underset{k=1}{\sum ^{m-1}}{k}^2+\underset{k=1}{\sum ^{m-1}}k+(n+1-m^2)m$$$$S_n=2\times \frac{(m-1)m(2m-1)}{6}+\frac{(m-1)m}{2}+(n+1-m^2)m$$$$S_n=\frac{(m-1)m(4m+1)}{6}+(n+1-m^2)m$$$$S_n=m(n+1)-\frac{m(m+1)(2m+1)}{6}$$$$withn=500:$$$$m=\lfloor \sqrt{500}\rfloor =22$$$$S_{500}=22\times 501-\frac{22\times 23\times 45}{6}=7227$$$$=====================$$$$\left[\sqrt{1}\right]+\left[\sqrt{2}\right]+\left[\sqrt{3}\right]+...+\left[\sqrt{n}\right]=m(n+1)-\frac{m(m+1)(2m+1)}{6}$$$$i.e.$$$$\left[\sqrt{1}\right]+\left[\sqrt{2}\right]+\left[\sqrt{3}\right]+...+\left[\sqrt{n}\right]=m(n+1)-\underset{k=1}{\sum ^m}{k}^2$$$$generally:$$$$\left[\sqrt[r]{1}\right]+\left[\sqrt[r]{2}\right]+\left[\sqrt[r]{3}\right]+...+\left[\sqrt[r]{n}\right]=m(n+1)-\underset{k=1}{\sum ^m}{k}^r$$$$withm=\lfloor \sqrt[r]{n}\rfloor$$

Commented by DAVONG last updated on 27/Mar/23

$$\mathrm{Thanks}\mathrm{sire}$$

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