Use technology, continual substitution over a system of this order leaves too much room for error.
0 12 10 0 -12 -20
I calculate the differences, and find it's a 3rd degree polynomial.
An^3+Bn^2+Cn+D
So I know when n=1 the above equation=0, and when n=2, 12, and so on.
So I come up with these 4 equations
a + b+ c+d=0
8a+ 4b+2c+d=12
27a+ 9b+3c+d=10
64a+16b+4c+d=0
Now I have to solve this system of linear equations. I just keep trying and getting wrong answers. What's the best way to attack this problem?
Since you have already calculated the differences, and seen that the "third differences" are all the same, I would suggest using "Newton's divided difference formula". The value and differences "at 0" are 0, 12, -14, 6 so the polynomial is
Check: , [itex]1^3- 10(1^2)+ 21(1)= 12[/tex], , , .