# Math Help - Next number in sequence problem/solving system of linear equations.

1. ## Next number in sequence problem/solving system of linear equations.

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?

2. ## Re: Next number in sequence problem/solving system of linear equations.

Use technology, continual substitution over a system of this order leaves too much room for error.

3. ## Re: Next number in sequence problem/solving system of linear equations.

Originally Posted by StudentMCCS
a + b+ c+d=0 [1]
8a+ 4b+2c+d=12 [2]
27a+ 9b+3c+d=10 [3]
64a+16b+4c+d=0 [4]
Do subtractions: [2]-[1], [3]-[1] and [4]-[1]; will leave
7a + 3b + c = 12 [5]
26a + 8b + 2c = 10 [6]
63a + 15b + 3c = 0 [7]

Divide [6] by 2 and [7] by 3:
7a + 3b + c = 12 [5]
13a + 4b + c = 5 [6]
21a + 5b + c = 0 [7]

Get it? Take over...

4. ## Re: Next number in sequence problem/solving system of linear equations.

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
$0+ 12x- (14/2!)x(x- 1)+ (6/3!)x(x-1)(x-2)= 12x- 7x(x- 1)+ x(x-1)(x-2)$ $= 12x- 7x^2+ 7x+ x^3- 3x^2+ 2x= x^3- 10x^2+ 21x$
Check: $0^3- 10(0^2)+ 21(0)= 0$, $1^3- 10(1^2)+ 21(1)= 12[/tex], $2^3- 10(2^2)+ 21(2)= 10$, [itex]3^3- 10(3^2)+ 21(3)= 0$, $4^3- 10(4^2)+ 21(4)= -12$, $5^3- 10(5^2)+ 21(5)= -20$.

5. ## Re: Next number in sequence problem/solving system of linear equations.

using matrices will be helpful for solving such equations

6. ## Re: Next number in sequence problem/solving system of linear equations.

Thanks everyone.