<s>[INST] Solve the linear homogeneous recurrence relation with constant coefficients $$9a_{n} = 6a_{n-1}-a_{n-2}, a_{0}=6, a_{1}=5$$
So
$$x^n = (6x^{n-1}-x^{n-2})\div9$$
thus
$$[x^2 = (6x-1)\div9] \equiv [x^2 - \frac{2}{3}x + \frac{1}{9} = 0], x=\frac{1}{3}$$
also
$$a_{2}=\frac{8}{3}, a_{3}=\frac{31}{27}$$
How do I plug in that x/root to solve for the given recurrence relation?
I tried
$$9a_{n} = 6(\frac{1}{3})^{n-1} - (\frac{1}{3})^{n-2}$$
which for n=1 gives 5... which is correct, for a_1 though, not 9a_1? So that's not right.
 [/INST] Since $a_n=c\left(\frac{1}{3}\right)^n+dn\left(\frac{1}{3}\right)^n$, $a_0=2\implies c=6$ and $a_1=5\implies d=9$, so
$\displaystyle a_n=\frac{6}{3^n}+\frac{9n}{3^n}=\frac{2+3n}{3^{n-1}}$
</s>