Solving a set of recurrence relation: Solving a set of recurrence relation: 
$a_n=3b_{n-1} , b_n=a_{n-1}-2b_{n-1}$
In addition, It's known that: $a_1=2, b_1=1$.
So i started by isolating $b_n \to b_n=3b_{n-2}-2b_{n-1}$
So i get the current equation: $x^2+2x-3=0 \to(x+3)(x-1)=0$, Which means $b(n)=A_1+A_2(-3)^n$
So what now? I found an equation with 2 parameters but only 1 relating b: $b_1=1$.
Can i use $a_1=2$ aswell on the found equation?
 A: Hint: You can calculate $b_2$ directly.
A: Define generating functions $A(z) = \sum_{n \ge 0} a_{n + 1} z^n$ and similarly $B(z)$. Write your recurrences as:
$$
\begin{align*}
a_{n + 1} &= 3 b_n \\
b_{n + 1} &= a_n - 2 b_n
\end{align*}
$$
By properties of generating functions, your recurrences translate to:
$$
\begin{align*}
\frac{A(z) - a_1}{z} &= 3 B(z) \\
\frac{B(z) - b_1}{z} &= A(z) - 2 B(z)
\end{align*}
$$
Thus:
$$
\begin{align*}
A(z) &= \frac{2 + 7 z}{1 + 2 z - 3 z^2} 
      = \frac{9}{4} \frac{1}{1 - z} - \frac{1}{4} \frac{1}{1 + 3 z} \\
B(z) &= \frac{1 + 2 z}{1 + 2 z - 3 z^2}
      = \frac{3}{4} \frac{1}{1 - z} + \frac{1}{4} \frac{1}{1 + 3 z}
\end{align*}
$$
Everything in sight is a geometric series:
$$
\begin{align*}
a_{n + 1} &= \frac{9}{4} - \frac{1}{4} (-3)^n \\
b_{b + 1} &= \frac{3}{4} + \frac{1}{4} (-3)^n
\end{align*}
$$
A: You are doing fine. You have 
$$ b(n)=A_1+A_2(-3)^n. $$
Now, to determine $A_1$ and $A_2$, you need two initial conditions $b_1$ and $b_2$. Already, you have got one which is $b_1$. To find $b_2$ use the second equation
$$ b_n=a_{n-1}-2b_{n-1} \implies b_2 = a_1-2b_1 =2-2=0 \implies b_2=0. $$
Now, you can advance.
