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Question Number 213413 by hardmath last updated on 04/Nov/24

$$\mathrm{Find}:$$$$\mathrm{A}=\left[\sqrt{1}\right]+\left[\sqrt{2}\right]+\left[\sqrt{3}\right]+...+\left[\sqrt{323}\right]=?$$

Answered by mehdee7396 last updated on 04/Nov/24

$$=(\left[\sqrt{1}\right]+\left[\sqrt{2}\right]+\left[\sqrt{3}\right]+)+(\left[4\right]+\left[5\right]+...+\left[\left[\sqrt{8}\right]\right)$$$$+(\left[\sqrt{9}\right]+\left[\sqrt{10}\right]+...+\left[\sqrt{24}\right])+...$$$$+(\left[\sqrt{289}\right]+\left[\sqrt{290}\right]+...+\left[\sqrt{323}\right])$$$$=3\left(1\right)+5\left(2\right)+7\left(3\right)+...+35\left(17\right)$$$$=\underset{n=1}{\sum ^{17}}(2n+1)n=2\underset{n=1}{\sum ^{17}}{n}^2+\underset{n=1}{\sum ^{17}}n$$$$=2\times \frac{17\times 18\times 35}{6}+\frac{17\times 18}{2}$$$$=3723$$$$$$

Commented by hardmath last updated on 04/Nov/24

$$$$dear friend, Is there an easier way to do this? I didn't quite understand it

Commented by hardmath last updated on 05/Nov/24

$$2\cdot \frac{17\cdot 18\cdot 35?}{6}\mathrm{dear}\mathrm{ser}\mathrm{how}?$$

Answered by Frix last updated on 05/Nov/24

$$\lfloor \sqrt{n}\rfloor =1,1⩽n⩽3\Rightarrow 3\times 1$$$$\lfloor \sqrt{n}\rfloor =2,4⩽n⩽8\Rightarrow 5\times 2$$$$\lfloor \sqrt{n}\rfloor =3,9⩽n⩽15\Rightarrow 7\times 3$$$$...$$$$\lfloor \sqrt{n}\rfloor =k,k^2⩽n⩽{(k+1)}^2-1=k^2+2k\Rightarrow (2k+1)\times k$$$$323={18}^2-1\Rightarrow 1⩽k⩽17$$$$\underset{k=1}{\sum ^m}(2k+1)k=2\underset{k=1}{\sum ^m}{k}^2+\underset{k=1}{\sum ^m}k=$$$$=\frac{m(m+1)(2m+1)}{3}+\frac{m(m+1)}{2}=$$$$=\frac{m(m+1)(4m+5)}{6}$$$$m=17\Rightarrow \mathrm{Answer}\mathrm{is}\frac{17\times 18\times 73}{6}=3723$$

Commented by hardmath last updated on 05/Nov/24

$$\mathrm{thankyoudear}\mathrm{professor}$$

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