Define p(i) as:
- 1, , when i = 1
- p(i - 1) + i * (i + 1) / 2 , when i > 1
Find p(1000)
- 1, , when i = 1
- p(i - 1) + i * (i + 1) / 2 , when i > 1
Find p(1000)
-
Hey mate! It's not as hard as it sounds.
The first six members of the sequence are 1, 4, 10, 20, 35 and 56. (I could go on but it's not needed.) The third-order differences are the same, so it's modeled by a cubic. Set up a 4-by-5 augmented matrix system with these rows: [1 1 1 1 1] [ 8 4 2 1 4] [27 9 3 1 10] [64 16 4 1 20] and use the matrix solving ability of a graphing calculator or Excel to solve. (I used OpenOffice and found the inverse matrix first, which I then multiplied by the "B" array.) This finds the cubic (1/6)x^3 + (1/2)x^2 + (1/3)x to model the sequence; evaluating for x = 1000 gives 167167000.
The first six members of the sequence are 1, 4, 10, 20, 35 and 56. (I could go on but it's not needed.) The third-order differences are the same, so it's modeled by a cubic. Set up a 4-by-5 augmented matrix system with these rows: [1 1 1 1 1] [ 8 4 2 1 4] [27 9 3 1 10] [64 16 4 1 20] and use the matrix solving ability of a graphing calculator or Excel to solve. (I used OpenOffice and found the inverse matrix first, which I then multiplied by the "B" array.) This finds the cubic (1/6)x^3 + (1/2)x^2 + (1/3)x to model the sequence; evaluating for x = 1000 gives 167167000.