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
$u_{n+k}=a^{k+1}u_{n-1}+b(a_k...(a_2(a_1+1)+1)...+1)$
 A: 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 $u_n=au_{n-1}+b$.
A: Just for illustration, a brute force approach yields
$$
\begin{split}
u_n &= au_{n-1} + b \\
    &= a(au_{n-2} + b) + b 
     = a^2 u_{n-2} + b(a+1)\\
    &= a^2 (au_{n-3} + b) + b(a+1)
     = a^3 u_{n-3} + b(a^2+a+1)\\
    &= a^nu_0 + b \sum_{k=0}^{n-1}a^k\\
    &= a^nu_0 + b \frac{1-a^n}{1-a} \\
    &= a^n \left( u_0-\frac{b}{1-a}\right) + \frac{b}{1-a}
\end{split}
$$
It remains to show that $$\frac{b}{1-a}$$ is a fixed point of the relation, to do that note that
$$
a\left(\frac{b}{1-a}\right) + b
= \frac{b}{1-a} \left[a + (1-a)\right]
= \frac{b}{1-a}
$$
A: Hint:  the fixed point property means that $d = ad+b\,$, then subtracting the two equations:
$$
\begin{align}
u_n &= a u_{n-1} + b \\
d &= ad + b
\end{align}
$$ 
gives $\,u_n - d = a(u_{n-1}-d)\,$. It follows by telescoping that:
$$u_n-d=a(u_{n-1}-d)=a^2(u_{n-2}-d)=\cdots=a^n(u_o-d)\,$$
