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The first four stellations are represented below. How many dots are in the 20th stellation?

 

Figure 1: one dot

Figure 2: eight dots

Figure 3: twenty-one dots

Figure 4: forty dots

 May 24, 2019
 #1
avatar+18937 
+1

             1

+7-0      8

+2*7-1   21

+3*7 -2  40

+4*7-3   65

+5*7-4  96

+6*7 - 5  133

+7*7 -6  176

+8*7 - 7 225

+9*7 -8  280

+10*7 -9  341

+11*7 -10   408     ??

 

Oops.....I thought it said 12th

20th would be

12*7 -11

13*7 -12

14*7 -13

15*7-14

16*7 -15

17*7 -16

18*7 -17

19*7 - 18   =   1160    ??

 May 24, 2019
edited by Guest  May 24, 2019
edited by ElectricPavlov  May 24, 2019
 #2
avatar+103097 
0

Very crafty, EP  !!!

 

 

cool cool cool

CPhill  May 24, 2019
edited by CPhill  May 24, 2019
 #3
avatar+18937 
+1

Crafty guesswork......trying to figure out a sigma notation for that mess-guess !  cheeky

ElectricPavlov  May 24, 2019
 #4
avatar+103097 
+1

EP's answer is more creative......but here's another way with the sum-of-differences

 

1         8            21            40  

 

      7        13             19

 

          6            6

 

We have two  non-zero rows....so.....this is a quadratic in the form  An^2 + Bn + C

 

So....we have this system of equations

 

A(1)^2  + B(1)  + C  = 1     ⇒  A + B + C  =  1    (1)

A(2)^2  + B(2) + C  = 8      ⇒ 4A + 2B + C  = 8   (2)

A(3)^2 + B(3)  +  C =  21   ⇒ 9A + 3B + C  = 21  (3)

 

Subtract (1) from (2)  and (2) from (3)   and we get

3A + B = 7     ⇒  -3A - B = -7   (4)

5A + B  = 13    (5)

 

Add  (4) and (5)  and we have that   2A  = 6    ⇒  A  = 3

And   3(3) + B = 7  ⇒  B = -2

And using (1)     3 - 2 + C  = 1  ⇒   C = 0

 

So....the generating polynomial  for the number of dots in the nth stellation   is    3n^2 - 2n

 

So....the number of dots in the 20th stellation is    3(20)^2 - 2(20)  = 1160

 

Which is the EXACT result that EP obtained !!!!

 

 

 

cool cool cool

 May 24, 2019
edited by CPhill  May 24, 2019
edited by CPhill  May 24, 2019
 #5
avatar+18937 
+2

I 'guess' this will work:

 

\( \sum_{1}^{20}\) (n-1)*7 - (n-2)    = 1160        Still rather clumsy....your answer is more elegant and 'math-y'  cheeky

ElectricPavlov  May 24, 2019
 #6
avatar+103097 
0

Mine might be more "math-y"......but it's also more " cook-book-y"

 

Your way required more ingenuity  [ IMHO  !!! ]

 

 

cool cool cool

CPhill  May 24, 2019
 #7
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0

Thank you Cphill and Electric Pavlov!

Guest May 24, 2019

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