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
+36916 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 ??
+36916 Crafty guesswork......trying to figure out a sigma notation for that mess-guess ! 
+128474 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 !!!!
+36916 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'
