Whats the best way to approach simplifying these three dependent relations:
$a_n = b_n + 2c_n$
$b_n = 2c_{n-1} $
$c_n = 2c_{n-1} + 2b_{n-1}$
(where $a_1=5, a_2=16$)
I tried to plug $b_n$ to $c_n$, and then simplifying $a_n$, but got stuck in the process. I know that the solution should be: $a_n = 2a_{n-1} + 4a_{n-2}$
if you take $$Y_n = \left( \begin{array}{c} b_n \\ c_n \end{array} \right) $$ then $$ Y_n = M Y_{n-1}, $$ where $$M = \left( \begin{array}{cc} 0&2 \\ 2&2 \end{array} \right) $$ I prefer writing $$ Y_{n+1} = M Y_{n}. $$ From the trace $2$ and determinant $-4,$ we find $M^2 - 2M-4I = 0,$ or $M^2 = 2M + 4I.$ This is Cayley-Hamilton. We point out that $$ Y_{n+2} = M^2 Y_{n}. $$ So $$ Y_{n+2} = M^2 Y_{n} = (2M+4I)Y_n = 2MY_n + 4 Y_n = 2 Y_{n+1} + 4 Y_n. $$ Or $$ Y_{n+2} = 2 Y_{n+1} + 4 Y_n. $$ Therefore $$ b_{n+2} = 2 b_{n+1} + 4 b_n $$ and $$ c_{n+2} = 2 c_{n+1} + 4 c_n. $$
What does this say about $a_n \; ?$
The first equation gives \begin{align} a_n & = b_n+2c_n\\ -2a_{n-1} & = -2b_{n-1}-4c_{n-1}\\ -4a_{n-2} & = -4b_{n-2}-8c_{n-2}. \end{align} Now the last two equations give $c_n=2c_{n-1}+4c_{n-2}$, by replacing $b_{n-1}$ in the last equation by using the second equation. Similarly we obtain $b_n=2b_{n-1}+4b_{n-2}$. Now summing up the three equations give the claim.
