Geometric/arithmetic sequences: $u_{n+1} = \frac12 u_{n} + 3$ I'm having trouble with writing this sequence as a function of $n$ because it's neither geometric nor arithmetic. 

$$\begin{cases}
u_{n+1} = \frac12 u_{n}  + 3\qquad \forall n \in \mathbb N\\
u_{0} = \frac13
\end{cases}$$

 A: Hint. Note that for some real number $a$ (which one?),
$$u_{n+1}-a = \frac12 \left(u_{n}-a\right).$$
Hence, the sequence $(u_n-a)_n$ is of geometric type:
$$u_{n}-a=\frac12 \left(u_{n-1}-a\right)=\frac{1}{2^2} \left(u_{n-2}-a\right)=\dots =\frac{1}{2^{n}} \left(u_{0}-a\right).$$
A: Similarly to what RobertZ did, turning the arithmetico-geometric recurrence to a purely geometric one, you can get rid of the multiplicative coefficient.
Let $u_n:=\dfrac{v_n}{2^n}$. Then
$$\dfrac{v_{n+1}}{2^{n+1}}=\frac12\dfrac{v_n}{2^n}+3$$ or
$$v_{n+1}=v_n+3\cdot2^{n+1}.$$
This recurrence is now easily solved as a geometric summation
$$v_n=6\,(2^n-1)+\frac13.$$
A: Write $$ 2u_{n+1}-u_n = 6 = 2u_n-u_{n-1}$$
so we have $$ 2u_{n+1}-3u_n +u_{n-1}=0$$
Then solving the characteristic equation $2x^2-3x+1=0$ which has a solution $x_1=1$ ans $x_2={1\over 2}$ we get a general solution $$ u_n = a\cdot  1^n +b\cdot \Big({1\over 2} \Big)^n$$
Since $u_1 = {19\over 6}$ by solving a system:
\begin{eqnarray*}
     {1\over 3} &=& a+b\\
 {19\over 6} &=& a+{b\over 2}
\end{eqnarray*}
we get the solution: 
 $$ u_n = 6-{17\over 3\cdot 2^n}$$
A: Let $u_m=v_m+a_0+a_1m+\cdots$
$$6=2u_{n+1}-u_n=2v_{m+1}-v_m+a_0(2-1)+a_1(2(m+1)-m)+\cdots$$
Set $a_0=6,a_r=0\forall r>0$
to find $$v_{n+1}=\dfrac{v_n}2=\cdots=\dfrac{v_{n-p}}{2^{p+1}}$$
Now $v_0+6=u_0\iff v_0=?$
