Question about recurrence relation problem.

Question

solve the following recurrence relation, subject to given initial conditions.

$a_{n+1} = 6a_n -9,$

$a_0 = 0,$

$a_1 = 3.$

Here is what I have done.

$a_{n+1} - 6a_n +9 = 0$

$a_n = r^n$

$r^{n+1} - 6r^n + 9 = 0$

$r + 6 = 0$

$r = -6$

$a_n$ = $X(-6)^n$

$a_1 = 3 = X(-6)^1$

$X = -1/2$

Therefore

$a_n = -1/2(-6)^n$

Am I on the right track or are there any mistakes I made somewhere?

Edited: added the capture of the problem

Answer 1

$$\begin{align} a_{n+1}&=6a_n-9\\ a_{n+1}-\frac95&=6a_n-\frac{54}5\\ &=6\left(a_n-\frac95\right)\\ u_{n+1}&=6u_n\\ u_n&=6u_{n-1}=6^2u_{n-2}=\cdots=6^{n-1}u_1\\ &=6^{n-1}\left(a_1-\frac95\right)\\ &=6^{n-1}\left(3-\frac95\right)\\ a_n-\frac95&=\frac65(6^{n-1})\\ &=\frac{6^n}5\\ a_n&=\frac15\left(9+6^n\right)\qquad\blacksquare \end{align}$$

Check: $$\begin{align} 6a_n-9&=\frac65(9+6^n)-9\\ &=\frac95+6(6^n)\\ &=\frac15(9+6^{n+1})\\ &=a_{n+1} \end{align} $$

Hence formula is correct.

Answer 2

This recurrence relation is nonhomogeneous, and cannot be solved directly using the characteristic polynomial (terms in the characteristic polynomial correspond to coefficients of $a_i$). Instead, try the following: \begin{align*} a_{n+1} &= 6a_n-9 \\ a_{n+2} &= 6a_{n+1}-9. \end{align*} Subtract the first equation from the second; this should give a second-order linear recurrence, which you can solve using the method above.