Solve the recurrence relation :

$\begin{aligned} &a_{n}+6 a_{n-1}+12 a_{n-2}+8 a_{n-3}=2^{n}, n \geq 3 \\\\ &a_{0}=0, a_{1}=0, a_{2}=2 \end{aligned}$

Solution:

The characteristic equation is,

$\begin{aligned} \Rightarrow \quad \lambda^{3}+6 \lambda^{2}+12 \lambda+8 &=0 \\\\(\lambda+2)^{3} &=0 \\\\ \quad \lambda =-2,-2,-2 \end{aligned} \\ $

Hence the homogeneous solution is,

$a_{n}^{(h)}=\left(A_{1} n^{2}+A_{2} n+A_{3}\right)(-2)^{n} \\ $

The particular solution will be of the form $P\left(2^{n}\right)$. Substituting the given recurrence relation, we get,

$ …