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

Guest May 24, 2019

#1**0 **

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 ??

ElectricPavlov May 24, 2019

#3**0 **

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

ElectricPavlov
May 24, 2019

#4**+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 !!!!

CPhill May 24, 2019

#5**0 **

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'

ElectricPavlov
May 24, 2019