I am trying to learn recurrence relations and I want to find the general formula of second-order linear homogeneous relation
$a_n=6a_{n-1}-8a_{n-2}$ where $a_0=1; a_1=4.$
So I started with the characteristic equation and I got this:
$r^2-6r+8=0$
After finding roots I should find a general formula for this relation:
$a_n=\alpha(4^n)+\beta(2^n)$
after plugging given values, I have got an equation:
$a_n=4^n$, and it's incorrect. Can someone help me and explain where I have a mistake? Also is it possible to solve this type of problem with the plug and chug method?
Thank you in advance
Best Answer
Plugging $n=0,1$ in the general solution $a_n = \alpha 4^n + \beta 2^n$ you get the linear system:
$$\begin{cases} \alpha + \beta = 1 \\ 4\alpha + 2\beta = 4\end{cases} \; ,$$
from which you obtain:
$$\alpha = 1,\ \beta = 0 \; .$$
Therefore your solution is correct... Why do you think it is not?