How would you prove this by induction? I've been trying to solve this one using induction for quite a while but I don't get to the solution. Any tips would be apreciated.
Let $a_1 = 47$, $a_2=80$ and for $n \geq 3$, $a_n =4a_{n-1} - 4a_{n-2}+3(n-2)^2$. Prove that for any positive integer $$ a_n=2^n(3+n)+3n^2+12n+24$$
 A: By Induction
Prove basis (trivial).
Induction step:
$\begin{align}
a_{n+1}=&4a_{n}-4a_{n-1}+3(n-1)^2 \\
=&4.2^n(3+n)+12n^2+48n+96 \\
-&4.2^{n-1}(2+n)-12(n-1)^2-48(n-1)-96 \\
+&3n^2-6n+3
\end{align}$,
by induction hypothesis.
$\implies a_{n+1}=2^{n+1}(3+n+1)+3(n+1)^2+12(n+1)+24$, by simplifying
$\therefore a_n=2^n(3+n)+3n^2+12n+24$, $\forall{n} \in \mathbb{N}$.
By solving the recurrence relation
Also, by solving the homogeneous recurrence relation $a_n=4a_{n-1}-4a_{n-2}$,
we have the characteristic equation $r^2-4r+4=(r-2)^2=0$,
hence $a_h(n)=c_12^n+nc_22^n$, for some constants $c_1, c_2$.
Also, guessing the particular solution $a_p(n)=An^2+Bn+C$ and by substitution, we have
$\begin{align}
An^2+Bn+C&=4A(n-1)^2+4B(n-1)+4C \\
&-4A(n-2)^2-4B(n-2)-4C \\ 
&+3(n^2-4n+4)
\end{align}$,
solving, we have $A=3$, $B=12$, $C=24$,
s.t., $a_p(n)=3n^2+12n+24$ and solution to the non-homogeneous recurrence relation is
$a_n=a_h+a_p=c_12^n+nc_22^n+3n^2+12n+24$,
now using boundary conditions $a_1=47$ and $a_2=80$, we have $c_1=3$, $c_2=1$
$\implies a_n=2^n(3+n)+3n^2+12n+24$
