proving recursively defined sequence by induction I would like to prove the following recursively defined sequence from $n-1$ to $n$ by induction. Im not realy sure about it. Any help or alternative ways to understand and prove it are highly appreciated :
$0,1,4,12,35,98$
$a_0=0$, $a_1=1$, $a_n=a_{n-1}+5a_{n-2}+3$ for  $n\geq2$ 
To prove $a_n\leq 3^n$
I thougt of it as: $a_{n-1}\leq 3^{n-1}$, $a_{n-2}\leq 3^{n-2}$ 
and thus:
$a_n\leq 3^{n-1} + 5\cdot 3^{n-2}+3$
$=3^{n-2} \cdot(3+5)+3$
$=3^{n-2} \cdot(8)+3$
$=3^{n-2} \cdot(9)+3$
$=3^{n-2} \cdot(3 \cdot 3)+3$
$=3^{n}+3$
$\leq 3^{n}$

It will be easier to show $a_n \le 3^n-1$ (and hence $3^n$).
$$\begin{align}
a_n &= a_{n-1} + 5a_{n-2}+3 \\ 
&\le (3^{n-1}-1) + 5(3^{n-2}-1) + 3 \\
&= 3^{n-1} + 5\cdot 3^{n-2} -3\\
&= 3^{n-1} + 3\cdot 3^{n-2} + 2\cdot 3^{n-2} -3\\
&= 3^{n-1} +  3^{n-1} + 2\cdot (3^{n-2} -1)-1\\
&\le 3^{n-1} +  3^{n-1} + 2\cdot 3^{n-2} - 1\\
&\le \underbrace{3^{n-1} +  3^{n-1} + 3^{n-1}}_{3^n} - 1
\end{align}$$
