I encountered this problem in the International baccalaureate Higher Level math book, and my teacher could not help. If anyone could please take the time to look at it, that would be great. I need help with question (b) in the attached picture. All of the relevant information is in there; I'm just conflicted about how to solve it.
Thanks :)
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For (b), the first recurrence implies $$ a_{n+2} = 3a_{n+1} + b_{n+1}. $$ Now substitute $a_{n+1} = 3a_n + b_n$ and $b_{n+1} = 5a_n - b_n$ into the right hand side of this equation and regroup to get $$ a_{n+2} = 3(3a_n + b_n) + 5a_n - b_n = 14a_n + 2b_n = 8a_n + 2(3a_n + b_n) = 8a_n + 2a_{n+1}. $$
Maybe try this: $$ a_{n+2} = 3a_{n+1}+b_{n+1}=9a_n+3b_n+4a_n-b_n=14a_n+2b_n \\ 2a_{n+1}+8a_n=6a_n+2b_n+8a_n=14a_n+2b_n $$ They are equal.