Find the general form of the solutions of the recurrence relations $a_n=8a_{n-2}-16a_{n-4}$

Question

Was hoping to confirm my steps and answer.

The characteristic equation of $a_n=8a_{n-2}-16a_{n-4}$ is $$r^4 = 8r^2 - 16$$ $$r^4 - 8r^2 + 16= r^4 - 4r^2 -4r^2 + 16= (r^2-4)(r^2-4) = (r-2)^2(r+2)^2$$ There are roots $2$ with multiplicity $2$ and $-2$ with multiplicity $2$ therefore solutions of this recurrence relation are of the form: $$a_n=\alpha_1 2^n+\alpha_2n2^n+\alpha_3(-2)^n+\alpha_4n(-2)^n$$

Answer 1

This looks good. You can check it with $$\begin{align}8a_{n-2}-16a_{n-4}=&\,2\alpha_12^{n}+2\alpha_2(n-2)2^{n}+2\alpha_3(-2)^{n}+2\alpha_4(n-2)(-2)^n\\&-\alpha_1 2^n-\alpha_2(n-4)2^n-\alpha_3(-2)^n-\alpha_4(n-4)(-2)^n\\=&\,a_n\end{align}$$ Also, since the difference equation only relates terms in the sequence of even separation, the odd and even terms evolve independently. So you can write, for odd $n$, $$a_n=\beta_12^n+\beta_2n2^n$$and for even $n$, $$a_n=\beta_32^n+\beta_4n2^n$$(Here, $\beta_1=\alpha_1-\alpha_3,\beta_2=\alpha_2-\alpha_4,\beta_3=\alpha_1+\alpha_3\,\beta_4=\alpha_2+\alpha_4$)

Answer 2

The characteristic equation you have got is actually the result of hoping a solution of the form $a_n=r^n$:

$r^n=8r^{n-2}-4r^{n-4}$ gives characteristic equation on simplification.