Recurrence relation:$$a_n = 6a_{n-1} – 9a_{n-2}$$
Initial conditions:$$a_1 = 1, a_2 = 9$$
I am having a bit of trouble finishing off this problem. So far I have:
Assume:$$a_n = r^n$$
$$r^n = 6r^{n-1}-9r^{n-2}$$
$$r^2 = 6r – 9$$
$$r^2 – 6r + 9 = 0$$
$$(r-3)^2 = 0$$
This means:
$$a_n = A(3)^n+B(n3^n)$$
$$a_1 = A3^1 + B((1)3^1) = 3A + 3B = 1$$
$$a_2 = A3^2 + B((2)3^2) = 9A + 18B = 9$$
From there I can't seem to combine the equations to find out what A and B are equal to correctly. Any help would be appreciated!
Best Answer
$$\begin{align} 9-3\times 1 &=a_2-3a_1 \\ 6 &= (9A+18B)-(9A+9B) \\ 6 &= 9B \end{align}$$ Therefore $B=\frac 23$, substitute this into $A_1$ yields $$ 3A=1-3B=1-3\times(\frac 23)=1-2=-1 $$ so that $A=-\frac 13$, thus $$ A=-\frac 13, B=\frac 23 $$