So I'm presented with no initial conditions and the following recurrence relation:
$8a_{n+2}+4a_{n+1}-4a_n=0$
I need to determine if these are possible solutions, and in that case, which initial conditions do they require:
- $a_n = 3(-1)^n$
- $a_n = 3(-1/2)^n + 1$
- $a_n = 7(1/2)^n$
I know this is a second degree linear recurrence so I've tried to reach a generic solution:
$8a_{n+2}+4a_{n+1}-4a_n=0$
$a_n=2a_{n+2}+a_{n+1}$
$r^n=c_1r^{n+1}+c_2r^{n+1}$
Dividing by $r^{n-2}$:
$r^2=rc_1 + c_2$
$0=-1r^2+2r+1$
$r_1=1-\sqrt{2}$
$r_2=1+\sqrt{2}$
$a_n=\alpha_1 r_{1}^{n} + \alpha_2 r_{2}^n$
But I don't know how to continue...
Hint: plug in what you are given $(a_{n})$ and see if it satisfies the recurrence. If it does, then compute $a_{0}$, $a_{1}$ and $a_{2}$ as your initial conditions. For example, if $a_{n}=3(-1)^{n}$ what we check is:
$8(3(-1)^{n+2})+4(3(-1)^{n+1})4(3(-1)^{n})=24(-1)^{n+2}+144(-1)^{2n+1}=-144\pm 24 \neq 0$ so the first case is not possible.
EDIT: Now that the recurrence has been edited to $8a_{n+2}+4a_{n+1}-4a_{n}=0$... we would have:
\begin{align*} 8(3(-1)^{n+2})+4(3(-1)^{n+1})-4(3(-1)^{n}) &= 24(-1)^{n+2}+4(3(-1)^{n+1})+4(-1)(3(-1)^{n}) \\ &= 24(-1)^{n+2}+12(-1)^{n+1}+12(-1)^{n+1} \\ &= 24(-1)^{n+2}+24(-1)^{n+1}=0 \end{align*}
so the $3(-1)^{n}$ does satisfy the recurrence and the initial conditions are $a_{0}=3(-1)^{0}=3$, $a_{1}=3(-1)^{1}=-3$ and $a_{2}=3$.