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 !!!!