Find [(√1)]+[(√2)]+[(√3)]+...+[(√(100))]=? with [x]=greatest integer function can we find a general formula for [(√1)]+[(√2)]+[(√3)]+...+[(√n)] in terms of n?


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Question Number 80804 by mr W last updated on 06/Feb/20

$F i n d$ $[\sqrt{1}] + [\sqrt{2}] + [\sqrt{3}] + . . . + [\sqrt{100}] = ?$ $w i t h [x] = g r e a t e s t i n t e g e r f u n c t i o n$ $c a n w e f i n d a g e n e r a l f o r m u l a f o r$ $[\sqrt{1}] + [\sqrt{2}] + [\sqrt{3}] + . . . + [\sqrt{n}]$ $i n t e r m s o f n ?$
Commented by MJS last updated on 06/Feb/20

$[\sqrt{k}] = n; n^{2} ⩽ k ⩽ n (n + 2)$ $does this help ?$
Commented by mr W last updated on 07/Feb/20

$y e s, i t h e l p s . t h a n k y o u s i r!$
	Answered by behi83417@gmail.com last updated on 07/Feb/20

	$[\sqrt{1}] + [\sqrt{2}] + [\sqrt{3}] = 3 \times 1$ $[\sqrt{4}] + . . . . . . . . . [\sqrt{8}] = 5 \times 2$ $[\sqrt{9}] + . . . . . . . . . [\sqrt{15}] = 7 \times 3$ $[\sqrt{16}] + . . . . . . . . [\sqrt{24}] = 9 \times 4$ $[\sqrt{25}] + . . . . . . . . [\sqrt{35}] = 11 \times 5$ $[\sqrt{36}] + . . . . . . . . [\sqrt{48}] = 13 \times 6$ $[\sqrt{49}] + . . . . . . . . . [\sqrt{63}] = 15 \times 7$ $[\sqrt{64}] + . . . . . . . . . [\sqrt{80}] = 17 \times 8$ $[\sqrt{81}] + . . . . . . . . . [\sqrt{100}] = 19 \times 9 + 10$ $Σ = 3 \times 1 + 5 \times 2 + 7 \times 3 + . . + 19 \times 9 + 10 = 625 ◼$
	Answered by mr W last updated on 07/Feb/20

	$S_{n} = [\sqrt{1}] + [\sqrt{2}] + [\sqrt{3}] + . . . + [\sqrt{n}]$ $l e t m = ⌊ \sqrt{n} ⌋$ $[\sqrt{1}] + [\sqrt{2}] + [\sqrt{3}] = 3 \times 1$ $[\sqrt{4}] + [\sqrt{5}] + [\sqrt{6}] + [\sqrt{7}] + [\sqrt{8}] = 5 \times 2$ $[\sqrt{9}] + [\sqrt{10}] + . . . + [\sqrt{15}] = 7 \times 3$ $. . .$ $[\sqrt{k^{2}}] + [\sqrt{k^{2} + 1}] + . . . + [\sqrt{k^{2} + 2 k}] = (2 k + 1) \times k$ $. . .$ $[\sqrt{m^{2}}] + [\sqrt{m^{2} + 1}] + . . . + \sqrt{n} = (n + 1 - m^{2}) \times m$ $S_{n} = \underset{k = 1}{\sum^{m - 1}} (2 k + 1) k + (n + 1 - m^{2}) m$ $S_{n} = 2 \underset{k = 1}{\sum^{m - 1}} k^{2} + \underset{k = 1}{\sum^{m - 1}} k + (n + 1 - m^{2}) m$ $S_{n} = 2 \times \frac{(m - 1) m (2 m - 1)}{6} + \frac{(m - 1) m}{2} + (n + 1 - m^{2}) m$ $S_{n} = \frac{(m - 1) m (4 m + 1)}{6} + (n + 1 - m^{2}) m$ $S_{n} = m (n + 1) - \frac{m (m + 1) (2 m + 1)}{6}$ $w i t h n = 100 :$ $m = ⌊ \sqrt{100} ⌋ = 10$ $S_{100} = 10 \times 101 - \frac{10 \times 11 \times 21}{6} = 625$ $w i t h n = 1000 :$ $m = ⌊ \sqrt{1000} ⌋ = 31$ $S_{1000} = 31 \times 1001 - \frac{31 \times 32 \times 63}{6} = 20615$ $=====================$ $w e s e e$ $[\sqrt{1}] + [\sqrt{2}] + [\sqrt{3}] + . . . + [\sqrt{n}] = m (n + 1) - \frac{m (m + 1) (2 m + 1)}{6}$ $i . e .$ $[\sqrt{1}] + [\sqrt{2}] + [\sqrt{3}] + . . . + [\sqrt{n}] = m (n + 1) - \underset{k = 1}{\sum^{m}} k^{2}$ $t h i s c a n b e g e n e r a l i s e d :$ $[\sqrt[r]{1}] + [\sqrt[r]{2}] + [\sqrt[r]{3}] + . . . + [\sqrt[r]{n}] = m (n + 1) - \underset{k = 1}{\sum^{m}} k^{r}$ $w i t h m = ⌊ \sqrt[r]{n} ⌋$ $e . g .$ $[\sqrt[5]{1}] + [\sqrt[5]{2}] + [\sqrt[5]{3}] + . . . + [\sqrt[5]{n}] = m (n + 1) - \underset{k = 1}{\sum^{m}} k^{5}$ $o r$ $[\sqrt[5]{1}] + [\sqrt[5]{2}] + [\sqrt[5]{3}] + . . . + [\sqrt[5]{n}] = m (n + 1) - \frac{m^{2} {(m + 1)}^{2} (2 m^{2} + 2 m - 1)}{12}$
	Commented by mr W last updated on 07/Feb/20

	$t h a n k y o u f o r t r y i n g a n d r e v i e w i n g s i r!$
	Commented by behi83417@gmail.com last updated on 07/Feb/20

	$thank you very much dear master .$ $it is wonderful and great .$
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