We can find a generating polynomial by the sum-of-differences technique
0, 6, 16, 30, 48
6 10 14 18
4 4 4
0 0
There are three differences required to get to the "0" row
This means that we have a possible 3rd degree polynomial with at least some of the coefficients ≠ 0
P(x) = ax3 + bx2 + cx + d
When x = 1, we have a + b + c + d = 0 (1)
When x = 2, we have 8a + 4b + 2c + d = 6 (2)
When x = 3, we have 27a + 9b + 3c + d = 16 (3)
When x = 4, we have 64a + 16b + 4c + d = 30 (4)
Subtracting (1) from (2) we have 7a + 3b + c = 6 (5)
Subtracting (2) from (3) we have 19a + 5b + c = 10 (6)
Subtracting (3) from (4) we have 37a + 7b + c = 14 (7)
Subtracting (5) from (6) we have 12a + 2b = 4 (8)
Subtracting (6) from (7) we have 18a + 2b = 4 (9)
And subtracting (8) from (9) we get
6a = 0 → a = 0
And back-substituting to find the other coefficients, we have
b= 2 → c = 0 → d = -2
And our polynomial turns out to be the quadratic
P(x) = 2x2 - 2
And the 40th term is given by
P(40 ) = 2(40)2 - 2 = 3198
We can find a generating polynomial by the sum-of-differences technique
0, 6, 16, 30, 48
6 10 14 18
4 4 4
0 0
There are three differences required to get to the "0" row
This means that we have a possible 3rd degree polynomial with at least some of the coefficients ≠ 0
P(x) = ax3 + bx2 + cx + d
When x = 1, we have a + b + c + d = 0 (1)
When x = 2, we have 8a + 4b + 2c + d = 6 (2)
When x = 3, we have 27a + 9b + 3c + d = 16 (3)
When x = 4, we have 64a + 16b + 4c + d = 30 (4)
Subtracting (1) from (2) we have 7a + 3b + c = 6 (5)
Subtracting (2) from (3) we have 19a + 5b + c = 10 (6)
Subtracting (3) from (4) we have 37a + 7b + c = 14 (7)
Subtracting (5) from (6) we have 12a + 2b = 4 (8)
Subtracting (6) from (7) we have 18a + 2b = 4 (9)
And subtracting (8) from (9) we get
6a = 0 → a = 0
And back-substituting to find the other coefficients, we have
b= 2 → c = 0 → d = -2
And our polynomial turns out to be the quadratic
P(x) = 2x2 - 2
And the 40th term is given by
P(40 ) = 2(40)2 - 2 = 3198