Solve recurrencerelation system
$$(1) a_1 = b_1 = c_1 = 1 $$ $$(2) a_{n-1} = a_n + b_n + c_n, n \ge 1 $$ $$(3) b_{n+1} = 4^n - c_n, n \ge 1 $$ $$(4) c_{n+1} = 4^n - b_n, n \ge 1 $$
Using generating function, at the end i got $$B(x) = C(x) = \frac{3x^2 - 4x^3}{(1-4x) (1+x)} = \frac{1/10}{(1-4x)}+ \frac{7/5}{(1+x)} $$
$$A(x) = \frac{B(x) + C(x) - 3}{x-1} = \frac{2(\frac{1/10}{(1-4x)}+ \frac{7/5}{(1+x)} - 3)}{x-1} $$
What should i do next? since we know that x-1 it can't be changed into generating function..
Define generating functions $A(z) = \sum_{n \ge 0} a_{n + 1} z^n$ and similarly $B(z), C(z)$. (the "$n + 1$" makes it overall more pleasant to work with by starting sums at 0). Shift recurrences by one (I believe it should be $a_{n + 1} = a_n + b_n + c_n$), multiply by $z^n$:
$\begin{align*} \sum_{n \ge 0} a_{n + 2} z^n &= \sum_{n \ge 0} a_{n + 1} z^n + \sum_{n \ge 0} b_{n + 1} z^n + \sum_{n \ge 0} c_{n + 1} z^n \\ \sum_{n \ge 0} b_{n + 2} z^n &= 4 \sum_{n \ge 0} 4^n z^n\sum_{n \ge 0} - c_{n + 1} z^n \\ \sum_{n \ge 0} c_{n + 2} z^n &= 4 \sum_{n \ge 0} 4^n z^n\sum_{n \ge 0} - b_{n + 1} z^n \end{align*}$
Recognize sums:
$\begin{align*} \frac{A(z) - a_1}{z} &= A(z) + B(z) + C(z) \\ \frac{B(z) - b_1}{z} &= \frac{4}{1 - 4 z} - C(z) \\ \frac{B(z) - b_1}{z} &= \frac{4}{1 - 4 z} - B(z) \end{align*}$
Use initial values, solve for the generating functions, as partial fractions:
$\begin{align*} A(z) &= \frac{1 - z - 4 z^2}{(1 - z) (1 + z) (1 - 4 z)} \\ &= \frac{2}{3 (1 - z)} - \frac{5}{1 + z} + \frac{8}{15(1 - 4 z)} \\ B(z) &= C(z) \\ &= \frac{1}{(1 + z) (1 - 4 z)} \\ &= \frac{5}{1 + z} + \frac{4}{5 (1 - 4 z)} \end{align*}$
Everything in sight is just geometric series:
$\begin{align*} a_n &= [z^{n - 1}] A(z) \\ &= \frac{2}{3} \cdot 1^{n - 1} - 5 \cdot (-1)^{n - 1} + \frac{8}{15} \cdot 4^{n - 1} \\ &= \frac{2}{3} + 5 \cdot (-1)^n + \frac{2}{15} \cdot 4^n \\ b_n &= c_n \\ &= [z^{n - 1}] B(z) \\ &= \frac{1}{5} \cdot 1^{n - 1} + \frac{4}{5} \cdot 4^{n - 1} \\ &= \frac{4^n - (-1)^n}{5} \end{align*}$