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$$\left\{\begin{aligned} a_n &&= &&2a_{n-1} + b_{n-1} + a_{n-2} - b_{n-2} && n \ge 2 && (1)\\ b_n &&=&& b_{n-1} + b_{n-2} - a_{n-2} && n \ge 2&& (2)\end{aligned}\right.$$

with $a_0 = 5, a_1 = 3, b_0 = 0, b_1 = 3$.

From (2) I get: $a_{n-2} = b_{n-1}+b_{n-2} - b_n$. Substituting in (1): $a_n = 2a_{n-1} + 2b_{n-1} - b_n$.

Now I'm stuck. I don't see what I can do next...

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Hint:

Adding the two equations gives $$ a_n + b_n = 2(a_{n-1} + b_{n-1}) \quad {\rm{for}} \quad n \ge 2 $$ so $$ a_n + b_n = 2^{n-1}(a_{1} + b_{1}) = 6 \cdot 2^{n-1} $$ Plugging this into the second recursion gives $$ b_n = b_{n-1} + 2 b_{n-2} - (a_{n-2} + b_{n-2}) = b_{n-1} + 2 b_{n-2} -6 \cdot 2^{n-3} $$

Likewise,the first equation gives $$ a_n = a_{n-1} + (a_{n-1}+ b_{n-1}) + 2 a_{n-2} - (a_{n-2} + b_{n-2}) = \\ = a_{n-1} + 6 \cdot 2^{n-2} + 2 a_{n-2} - 6 \cdot 2^{n-3} = \\ = a_{n-1} + 2 a_{n-2}+ 3 \cdot 2^{n-2} $$

So there are two single-variable recursions which can be solved.

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  • $\begingroup$ Good. I introduced $c_n= a_{n-1}, \; d_n = b_{n-1},$ wrote it all as a 4 by 4 $X_{n+1} = MX_n,$ then revised finding the Jordan form of $M.$ After many pages, i got $a_n = (n+4)2^{n-1} + 2(-1)^n$ $\endgroup$ – Will Jagy Jan 29 at 18:42

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