First subtract the first term from each of the others to give us a series with three terms:
0, 3, 9.
We are told that this is a quadratic series, so let f(x)=ax²+bx where a and b are constants.
f(1)=0, f(2)=3, f(3)=9.
We get three equations:
f(1)=a+b=0, so b=-a; f(2)=4a+2b=3; f(3)=9a+3b=9. But 9a+3b=9 can be written 3a+b=3.
We only need two equations to find two unknowns, and so let’s see what happens if we substitute b=-a in f(2) and f(3):
4a-2a=3, so 2a=3 and a=3/2; 3a-a=3, so 2a=3 and a=3/2. These two equations are consistent so a=3/2 and b=-a=-3/2 is the solution. f(x)=3x²/2-3x/2.
The series is generated using the formula 3x²/2-3x/2+6, because we need to add back the first term and if we start with x=0 the formula gives us 6, x=1 gives us 6, x=2 gives us 9, x=3 gives us 15.