<s>[INST] 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
$u_{n+k}=a^{k+1}u_{n-1}+b(a_k...(a_2(a_1+1)+1)...+1)$
 [/INST] 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
$$
a\left(\frac{b}{1-a}\right) + b
= \frac{b}{1-a} \left[a + (1-a)\right]
= \frac{b}{1-a}
$$
</s>