<s>[INST] Solving a recurrence relation without the characteristic equations I am new to solving recurrence relations and I am presented with the following two problems 
(1) $$a_n = (3n-1)a_{n-1}, a_0 = 3$$ 
and
(2) $$a_{n+3}-6a_{n+2} + 11a_{n+1}-6a_{n} = 0, a_0 =1, a_1 =1, a_2 = -2$$
For (1), I try to find a pattern so $a_n = (3n-1)(3(n-1)-1)a_{n-2} = (3n-1)(3n-4)a_{n-2}$ Now do this one more time so we have $a_n = (3n-1)(3n-4)(3n-7)a_{n-4}$ and so we see a pattern emerging
So I just conclude that $$a_n =(3n-1)(3n-4)(3n-7) \dots \times 3$$ But I feel there is something wrong here.
For (2) Let $a_n = cr^n$ where $c,r \neq 0$and then I use the characteristic equation $$r^3 -6r^2 + 11r - 6 = 0 =(r-3)(r-2)(r-1)=0$$
So I have $r=1,2,3$ as the characteristic roots
Now we have $a_n = A(1)^n + B(2)^n + C(3)^n$
$1 = A + B + C$
$1 = A + 2B + 3C$
$-2 = A + 4B + 9C$
And after solving this linear system, I get $$A= \frac{-1}{2}, B=3, C=\frac{-3}{2}$$
And so finally we have that $$a_n = \frac{-1}{2} + 3(2)^n  -\frac{3^{n+1}}{2}, n \geq 0$$
I think I got the second recurrence right, But the first one I don't know how to find it
 [/INST] $$a_n = (3n-1)a_{n-1}=(3n-1)(3n-4)a_{n-2}=...=a_{n-m}\prod_{k=1}^m(3n-3k+2)\\
=a_{n-n}\prod_{k=1}^n(3n-3k+2)=a_0\prod_{k=1}^n(3n-3k+2)$$
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