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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.

enter image description here

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

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

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