Help solving recurrence relation, $a_n = 3a_{n-1} + 4a_{n-2} - 12a_{n-3}$

Question

This is in my homework, and I am not sure how to go about this, I've read the book but I can't seem to grasp what to do. Help?

$$a_n = 3a_{n-1} + 4a_{n-2} - 12a_{n-3}$$

where $a_0 = 2$, $a_1 = -1$, $a_2 = 3$

Answer 1

Hint: the characteristic polynomial $x^3-3x^2-4x+12$ factors as $(x-3)(x-2)(x+2)$.

Answer 2

Use the techniques in Wilf's "generatingfunctionology". Write: $$ a_{n + 3} = 3 a_{n + 2} + 4 a_{n + 1} - 12 a_n $$ Define $A(z) = \sum_{n \ge 0} a_n z^n$, multiply the recurrence by $z^n$ and sum over $n \ge 0$ to get: $$ \frac{A(z) - a_0 - a_1 z - a_2 z^2}{z^3} = 3 \frac{A(z) - a_0 - a_1 z}{z^2} + 4 \frac{A(z) - a_0}{z} - 12 A(z) $$ Solve for $A(z)$, split into partial fractions, and read off the coefficients.