Getting rid of exponents with n when solving with annihilators: $a_n=a_{n-1}+2a_{n-2}+2^n+n^2$

Question

To solve the following with annihilators:

$a_n=a_{n-1}+2a_{n-2}+2^n+n^2$, for $n\ge2$, with initial conditions $a_1=0$ and $a_0=0$

we would have to get rid of the $2^n$ term at least, otherwise any E<> operation will just reduce the exponent. If we try to get the $2^n$ term to cancel out somehow by picking a convenient substitute, then the $n^2$ term becomes problematic. Dividing through makes the equation too complicated for annihilators.

Answer 1

First let $a_n = b_n + xn^2 + yn +z$. Compute $x,y,z$ such that we have $$b_n = b_{n-1} + 2b_{n-2} + 2^n$$ Now set $c_n = b_n/2^n$, we then have $$c_n = \dfrac{c_{n-1}}2 + \dfrac{c_{n-2}}2 + 1$$ Now you should be able to take it from here.