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Question Number 55491 by Tinkutara last updated on 25/Feb/19

	Answered by tanmay.chaudhury50@gmail.com last updated on 26/Feb/19
	$trying to understand restricting under following condition 1)radius(r) taken as integer 2)centre and otherpoints lie on the circle not counted in f(r) discussion... let r=1 then intregal points are [(0,0),(1,0),(0,1),(0,−1),(−1,0)] but these red marked points not considered so f(1)=0 r=2 [(0,0),(1,0),(−1,0),(0,1),(0,−1),(1,1),(1,−1),(−1,1),(−1,−1)] x={−1,0,1} y={−1,0,1} 3c_1 ×3c_1 −1=8 r=3 x={−2,−1,0,1,2} y={−2,−1,0,1,2} 5c_1 ×5c_1 −1=24 r=4 x={−3,−2,−1,0,1,2,3} y={−3,−2,−1,0,1,2,3} 7c_1 ×7c_1 −1=49−1 [srl no] [r ] [nos of intregal points] [f(r)] 1. 1 [null] f(1)=0 2. 2 [8] f(2)=8 3 3 [24] f(3)=24 ..... .... f(r) =(1−1)+(9−1)+(25−1)+(49−1)+... =[(1^2 +3^2 +5^2 +7^2 +...(2r−1)^2 ]−r others pls check is it correct perception.. then i shall proceed further... =((r(2r+1)(2r−1))/3)−r =((r(4r^2 −1)−3r)/3) =((4r^3 −r−3r)/3) =((4r^3 −4r)/3) li_(r→∞)$
	$t r y i n g t o u n d e r s t a n d$ $r e s t r i c t i n g u n d e r f o l l o w i n g c o n d i t i o n$ $1) r a d i u s (r) t a k e n a s i n t e g e r$ $2) c e n t r e a n d o t h e r p o i n t s l i e o n t h e c i r c l e$ $n o t c o u n t e d i n f (r)$ $d i s c u s s i o n . . .$ $l e t r = 1 t h e n i n t r e g a l p o i n t s a r e [(0, 0), (1, 0), (0, 1), (0, - 1), (- 1, 0)]$ $b u t t h e s e r e d m a r k e d p o i n t s n o t c o n s i d e r e d$ $s o f (1) = 0$ $r = 2 [(0, 0), (1, 0), (- 1, 0), (0, 1), (0, - 1), (1, 1), (1, - 1), (- 1, 1), (- 1, - 1)]$ $x = {- 1, 0, 1} y = {- 1, 0, 1} 3 c_{1} \times 3 c_{1} - 1 = 8$ $r = 3 x = {- 2, - 1, 0, 1, 2} y = {- 2, - 1, 0, 1, 2}$ $5 c_{1} \times 5 c_{1} - 1 = 24$ $r = 4 x = {- 3, - 2, - 1, 0, 1, 2, 3}$ $y = {- 3, - 2, - 1, 0, 1, 2, 3}$ $7 c_{1} \times 7 c_{1} - 1 = 49 - 1$ $[s r l n o] [r] [n o s o f i n t r e g a l p o i n t s] [f (r)]$ $1 . 1 [n u l l] f (1) = 0$ $2 . 2 [8] f (2) = 8$ $3 3 [24] f (3) = 24$ $. . . . .$ $. . . .$ $f (r)$ $= (1 - 1) + (9 - 1) + (25 - 1) + (49 - 1) + . . .$ $= [(1^{2} + 3^{2} + 5^{2} + 7^{2} + . . . {(2 r - 1)}^{2}] - r$ $o t h e r s p l s c h e c k i s i t c o r r e c t p e r c e p t i o n . .$ $t h e n i s h a l l p r o c e e d f u r t h e r . . .$ $= \frac{r (2 r + 1) (2 r - 1)}{3} - r$ $= \frac{r (4 r^{2} - 1) - 3 r}{3}$ $= \frac{4 r^{3} - r - 3 r}{3}$ $= \frac{4 r^{3} - 4 r}{3}$ $\underset{r \to \infty}{li}$
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