# Solve the recurrence relation $a_n=6a_{n-1}-9a_{n-2}-8$ for $n\geq2$, $a_0=0$, $a_1=1$

My task: $$a_n=6a_{n-1}-9a_{n-2}-8$$ for $$n\geq2$$, $$a_0=0$$, $$a_1=1$$

My solution $$x^{2}-6x+9$$

$$\Delta=0$$

$$x_0=3$$

So I am gonna use following formula: $$a_n=ar^{n}+bnr^{n}$$

$$a_n=a*(3)^{n}+bn*3^{n}$$

$$-8$$ is the problem, so I am looking for $$c$$ that $$b_n:=a_n+c\implies b_n=6b_{n-1}-9b_{n-2}$$

$$b_n=6(b_{n-1}-c)-9(b_{n-2}-c)-8+c=6b_{n-1}-9b_{n-2}-8+6c$$ I am setting $$c=\frac{4}{3}$$ so

$$b_n=6b_{n-1}-9b_{n-2}\implies\exists a,\,b:\,b_n=a*3^{n}+bn*3^{n}.$$From $$b_0=\frac{4}{3},\,b_1=\frac{7}{3}$$, after finding $$a,\,b$$. Then $$a_n=b_n-\frac{1}{2}$$.

$$a=\frac{4}{3}$$ $$b=-\frac{5}{9}$$ $$b_2=22$$

$$a_2=22-\frac{4}{3}=\frac{62}{3}$$

Actual $$a_2=-2$$

So $$a_2$$ from $$b_n$$ method is not equal to actual $$a_n$$.

Can I use this $$b_n$$ method if delta equals 0? or should $$c=-\frac{4}{3}$$?

• How did you compute $b_0$ and $b_1$? – Exodd Jan 27 at 21:03
• $$b_0=a_0+\frac{4}{3}=\frac{4}{3}$$ $$b_1=a_1+\frac{4}{3}=\frac{7}{3}$$ Forgot to edit – Arnolt Infern Kitler Jan 27 at 21:09

I think that your choice of $$c$$ is wrong From $$b_n=6(b_{n-1}-c)-9(b_{n-2}-c)-8+c=6b_{n-1}-9b_{n-2}-8+4c$$
this gives you $$c=2$$ and $$b_n=a_n+2$$
$$b_0=2, b_1=3$$ after solving $$b_n=a*3^{n}+bn*3^{n}$$ you get $$a=2 ,b=-1$$