Recurrence relation $a_n = 11a_{n-1} - 40a_{n-2} + 48a_{n-3} + n2^n$ I didn't do a lot of maths in my career, and we asked me to solve the following recurrence relation:
$$a_{n} = 11a_{n-1} - 40a_{n-2} + 48a_{n-3} + n2^n$$
with 
$a_0 = 2$, $a_1 = 3$ and $a_2 = 1$
What is the procedure to solve such relation? So far, I just know that $a_n$ could be split into two difference recurrence relations (homogeneous and non-homogeneous) as $a_n = b_n + c_n$, where 
$$b_n = 11b_{n-1} - 40b_{n-2} + 48b_{n-3}$$
and 
$$c_n = n2^n$$

You can try to find coefficients $k_{1}$ and $k_{2}$ so that taking $b_{n}=a_{n}-k_{1}a_{n-1}+k_{2}a_{n-2}$, we might have an equation of the form $b_{n}-\alpha b_{n-1}=n2^{n}$, for some $\alpha$. You can work it out and one possibility is $\alpha=4$, $b_{n}=a_{n}-7a_{n-1}+12a_{n-2}$(The root of the equation $x^{3}-11x^{2}+40x-48=0$ being 3,4,4). So you need to solve $b_{n}-4b_{n-1}=n.2^{n}$. You can find $b_{n}$ through telescoping.
After this, knowing $b_{n}$, taking $d_{n}=a_{n}-3a_{n-1}$, we get $d_{n}-4d_{n-1}=b_{n}$, and then you get  $ d_{n}$ (knowing $b_{n}$) through telescoping, and then you finally get $a_{n}$ (now knowing $d_{n}$) again through telescoping.
