# Partial fraction of a generating function

I am solving a recurrence relation

$a_0 = a_1 = a_2 = 1, a_{n+3} = a_{n+2} − 2a_{n+1} − 4a_n$ for $n \ge 0$

I got a generating function for this sequence

$f(x) = \frac{2x^2+1}{4x^3+2x^2-x+1}$

Now I want to get the formula for the n-th term of this sequence. And so I wanted to use the partial fractions method but so far I was unsuccesful in finding the partial fraction. I'm probably missing something simple here.

• $x=-1$ cancels the denominator. The other roots are complex.. – Claude Leibovici Apr 2 '16 at 8:26

## 1 Answer

hint:

$$4x^3 +2x^2 -x +1=(4x^2 -2x +1) (x+1)$$

• so now I got the partial fraction of $f(x) = \frac{3}{7(x+1)} + f(x) = \frac{2x+4}{7(4x^2-2x+1)}$ How can I manipulate the second fraction with x in the numerator to get the n-th term formula? The first fraction is easy but I'm not so sure about the second one @MotylaNogaTomkaMazura – mathew7k5b Apr 2 '16 at 9:03
• $$\frac{2x+4}{4x^2 -2x +1} =\frac{(2x+4)(2x+1)}{(2x)^3 +1} =\frac{(2x+4)(2x+1)}{1+(2x)^3 }=\frac{(2x)^2}{1+(2x)^3 } +\frac{5\cdot (2x)}{1+(2x)^3 }+$$ $$\frac{4}{1+(2x)^3 } =(2x)^2 \sum_{n=0}^{\infty} (-(2x)^3)^n+5\cdot (2x)\sum_{n=0}^{\infty} (-(2x)^3)^n +4 \sum_{n=0}^{\infty} (-(2x)^3)^n$$ – MotylaNogaTomkaMazura Apr 2 '16 at 11:06