Solving this inhomogenous system of two linear recurrence relations I have this system of recurrence relations (where $\forall n\in\Bbb N: b^{(n)}, c^{(n)}\in\Bbb R$):
$$
\begin{pmatrix}b^{(n+1)}\\c^{(n+1)}\end{pmatrix} = \begin{pmatrix}1\over2^{n+1}\\0\end{pmatrix} + \begin{pmatrix}\frac34 & \frac14 \\ \frac14 & \frac34\end{pmatrix} \begin{pmatrix}b^{(n)} \\ c^{(n)}\end{pmatrix}.
$$
How can I find an explicit formula for all the $b^{(n)}, c^{(n)}$?
My question stems from this question. The latter question is trivial to solve if my question from above is answered. 
I thought about using some fixed point Theorem, but the calculations seem to get messy.
 A: Hint.
Making
$$
M=\frac 14\left(
\begin{array}{cc}
 3 & 1 \\
 1 & 3 \\
\end{array}
\right) = V^{\dagger}\cdot\Lambda\cdot V,\ \ U_n=\left(
\begin{array}{c}
 b_n \\
 c_n \\
\end{array}
\right)
$$
with 
$$
V = \frac{1}{\sqrt 2}\left(
\begin{array}{cc}
 1 & 1 \\
 -1 & 1 \\
\end{array}
\right)
$$
and 
$$
\Lambda = \left(
\begin{array}{cc}
 1 & 0 \\
 0 & \frac 12 \\
\end{array}
\right)
$$
Then
$$
U_{n+1}=V^{\dagger}\cdot \Lambda\cdot V\cdot U_n + \delta_n
$$
This is a linear recurrence with solution
$$
U_n = U_n^h+U_n^p\\
U_{n+1}^h - M\cdot U_n^h = 0\\
U_{n+1}^p - M\cdot U_n^p =\delta_n
$$
then
$$
U_n^h = V^{\dagger}\cdot\Lambda^n\cdot V\cdot C_0
$$
now calling $U_n^p = V^{\dagger}\cdot \Lambda\cdot V\cdot C_n$ and substituting we have
$$
V^{\dagger}\cdot\Lambda^{n+1}\cdot VC_{n+1}=V^{\dagger}\cdot\Lambda^{n+1}\cdot VC_n +\delta_n
$$
giving
$$
C_{n+1}-C_n  = V\Lambda^{-(n+1)}\cdot V^{\dagger}\cdot \delta_n = \frac 14\left(
\begin{array}{c}
2+2^{-n} \\
-2+2^{-n} \\
\end{array}
\right)
$$
etc.
and finally
$$
U_n = V^{\dagger}\cdot\Lambda^n\cdot V\cdot\left(C_0+C_n\right)
$$
NOTE
According to @G Cab calling $W = V\cdot U$ and $\rho = V\cdot\delta_n$ we can write also
$$
W_{n+1} = \Lambda\cdot W_n + \rho_n
$$
which is easier to handle.
