Question: I'm having trouble with writing this sequence as a function of $n$ because it's neither geometric nor arithmetic. 

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

Answer: 
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:
$$u\_{n}-a=\frac12 \left(u\_{n-1}-a\right)=\frac{1}{2^2} \left(u\_{n-2}-a\right)=\dots =\frac{1}{2^{n}} \left(u\_{0}-a\right).$$
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 $v\_0+6=u\_0\iff v\_0=?$