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