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-

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

F i n d

[\sqrt{1}] + [\sqrt{2}] + [\sqrt{3}] + \dots + [\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}] + \dots + [\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}] + \dots \dots \dots [\sqrt{8}] = 5 \times 2

[\sqrt{9}] + \dots \dots \dots [\sqrt{15}] = 7 \times 3

[\sqrt{16}] + \dots \dots . . [\sqrt{24}] = 9 \times 4

[\sqrt{25}] + \dots \dots . . [\sqrt{35}] = 11 \times 5

[\sqrt{36}] + \dots \dots . . [\sqrt{48}] = 13 \times 6

[\sqrt{49}] + \dots \dots \dots [\sqrt{63}] = 15 \times 7

[\sqrt{64}] + \dots \dots \dots [\sqrt{80}] = 17 \times 8

[\sqrt{81}] + \dots \dots \dots [\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}] + \dots + [\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}] + \dots + [\sqrt{15}] = 7 \times 3

\dots

[\sqrt{k^{2}}] + [\sqrt{k^{2} + 1}] + \dots + [\sqrt{k^{2} + 2 k}] = (2 k + 1) \times k

\dots

[\sqrt{m^{2}}] + [\sqrt{m^{2} + 1}] + \dots + \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}] + \dots + [\sqrt{n}] = m (n + 1) - \frac{m (m + 1) (2 m + 1)}{6}

i . e .

[\sqrt{1}] + [\sqrt{2}] + [\sqrt{3}] + \dots + [\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}] + \dots + [\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}] + \dots + [\sqrt[5]{n}] = m (n + 1) - \underset{k = 1}{\sum^{m}} k^{5}

o r

[\sqrt[5]{1}] + [\sqrt[5]{2}] + [\sqrt[5]{3}] + \dots + [\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 .