# Solving Recurrence Relation $a_n=6a_{n-1} - 9a_{n-2}$ for $n\geqslant2$

So the problem is to solve this recurrence relation with the initial conditions $a_0 = 2, a_1 = 21$.

$a_n=6a_{n-1} - 9a_{n-2}$ for $n\geqslant2$

And also find the value of $a_{2016}.$

Here is my solution but I'm not entirely sure if it's correct. Was wondering if anyone can confirm what I did is valid or perhaps I made a mistake somewhere? Thanks in advance.

• Sounds about right. – Ivan Neretin Apr 12 '17 at 21:44
• @IvanNeretin thanks! – Jamie Apr 12 '17 at 21:46
• Very good. continue. – hamam_Abdallah Apr 12 '17 at 22:11
• @Jamie here is an other method. – hamam_Abdallah Apr 12 '17 at 22:26

Put $a_n=3^nb_n$.

then

$$3^nb_n=2.3^nb_{n-1}-3^nb_{n-2}$$

or

$b_n-b_{n-1}=b_{n-1}-b_{n-2}$ =constante $$=b_1-b_0=7-2=5$$

thus

$$b_n=5n+2$$

and

$$\boxed {a_n=3^n (5n+2)}$$