Closed form of $a_{k+2} = 4a_k + c^2$ when $c$ is some constant.

Question

What is the closed form of $a_{k+2} = 4a_k + c^2$ when $c$ is some constant.

How can we find the closed form of this recurrence with constant? Usually , I’ll use the characteristic root technique, but now there’s some constant. Thanks in advance!

Answer 1

Giben $$a_{n+2}-4a_n=c^2~~~~~(1)$$ Let us first solve $a_{n+2}=4a_{n} ~~~~(2) \implies x^2=4 \implies x=\pm 2$. So the solution of (2) is $a_n= C_1 2^n +C_2 (-2)^n$ For (1) let $a_n=A$, we get $A=-c^2/3$. So the soluttion of (1) is $$a_n=C_1 2^n+ C_2~ (-2)^{n}-c^2/3$$