Given $u_n=au_{n-1}+b$ show that $u_n=A\cdot a^n + d$ where $d$ is a fixed point of $u_n$ Given $u_n=au_{n-1}+b$ show that $u_n=A\cdot a^n + d$ where:


*

*$d$ is a fixed point of the recurrence relation $u_n=au_{n-1}+b$;

*$A$ is a constant.


My attempt:
I have tried evaluating the recurrence relation and I've come to the following equation
$u_{n+k}=a^{k+1}u_{n-1}+b(a_k...(a_2(a_1+1)+1)...+1)$

We have 
$$\frac{u_n}{a^n}=\frac{u_{n-1}}{a^{n-1}}+\frac{b}{a^n}$$
So,
\begin{align}
\sum_{k=1}^n\frac{u_k}{a^k}&=\sum_{k=1}^n\frac{u_{k-1}}{a^{k-1}}+\sum_{k=1}^n\frac{b}{a^k}\\
\frac{u_n}{a^n}&=u_0+\frac{\frac{b}{a}\left(1-\left(\frac{1}{a}\right)^n\right)}{1-\frac{1}{a}}\\
&=u_0+\frac{b\left(1-\left(\frac{1}{a}\right)^n\right)}{a-1}\\
u_n&=\left(u_0+\frac{b}{a-1}\right)a^n-\frac{b}{a-1}
\end{align}
If $\displaystyle u_0=-\frac{b}{a-1}$, then $\displaystyle u_n=-\frac{b}{a-1}$ for all $n\in\mathbb{N}$.
$\displaystyle -\frac{b}{a-1}$ is a fixed point of the recurrence relation $u_n=au_{n-1}+b$.
