From the following recurrence relation:
$a_n =- a_{n-1}+8a_{n-2}+12a_{n-3}+25\cdot3^{n-2}-18n^2+48n+14$, for $n\geq3$
Where $a_0=6, a_1 = 0 $ and $a_2=57$.
My attempt:
I have generated a sequence but I have found that the result is simply too large to put into a form that can be decomposed into partial fractions.
My result is as follows:
Any help or hints on how I can go about this are much appreciated.
The recurrence relation seems to have disappeared from the question, so I repeat it here: $a_n=-a_{n-1}+8a_{n-2}+12a_{n-3}+25\cdot3^{n-2}-18n^2+48n+14$, with $a_0=6,a_1=0,a_2=57$.
Ignoring the last four terms, we have a standard linear recurrence relation. The associated polynomial has roots 3,-2, so its general solution is $A3^n+B(-2)^n+Cn(-2)^n$.
Next we need to find a particular solution. For the $25\cdot3^{n-2}$ we expect $Dn3^n$ to work for some $D$ (since $3^n$ is part of the general solution). For the $-18n^2+48n-14$ we expect $En^2+Fn+G$ to work. We duly find $n3^n+n^2+3n$.
So finally we have to get $A,B,C$ from the given values for $a_0,a_1,a_2$, giving $A=1,B=5,C=0$, hence $a_n=3^n+5(-2)^n+n 3^n+n^2+3n$.