How to find the solution of integer equation group

Question

I have the following problem: to find the general item of the following equation: let $a_1=b_1=1$, $$a_{n+1}=6a_n+2b_n, b_{n+1}=3a_n+2b_n$$ for any $n\geq 1$, find $a_n=?, b_n=?$

Answer 1

Since $$a_{n+1}=6a_n+2b_n\Rightarrow b_n=\frac{a_{n+1}-6a_n}{2}\tag1$$ we have $$b_{n+1}=3a_n+2b_n\Rightarrow \frac{a_{n+2}-6a_{n+1}}{2}=3a_n+(a_{n+1}-6a_n)\Rightarrow a_{n+2}-8a_{n+1}+6a_n=0.$$ So, letting $\alpha\lt\beta$ be the roots of $x^2-8x+6=0$, we have $$a_{n+1}-\alpha a_n=\beta(a_n-\alpha a_{n-1})=\cdots=\beta^{n-1}(a_2-\alpha a_1)$$ $$a_{n+1}-\beta a_n=\alpha(a_n-\beta a_{n-1})=\cdots=\alpha^{n-1}(a_2-\beta a_1)$$ Subtracting the latter from the former gives you $$\begin{align}(\beta-\alpha)a_n&=(\beta^{n-1}-\alpha^{n-1})a_2+(\alpha^{n-1}\beta-\alpha\beta^{n-1})a_1\\&=8(\beta^{n-1}-\alpha^{n-1})+\alpha\beta(\alpha^{n-2}-\beta^{n-2})\\&=8(\beta^{n-1}-\alpha^{n-1})+6(\alpha^{n-2}-\beta^{n-2}).\end{align}$$ Hence, we have $$a_n=\frac{4\left((4+\sqrt{10})^{n-1}-(4-\sqrt{10})^{n-1}\right)+3\left((4-\sqrt{10})^{n-2}-(4+\sqrt{10})^{n-2}\right)}{\sqrt{10}}.$$ You can get $b_n$ from $(1)$.