Solving a finite linear recurrence relation of the form $s_{i+1}=\alpha s_i + \beta$

Question

Suppose I have a sequence $s_0, s_1,...,s_N$, where $N$ is a finite number. This sequence obeys the recurrence relation $s_{i+1}=\alpha s_i+\beta$. The sequence also satisfies the following conditions: $s_0=0$, $s_1=r_{min}$ , and $s_N=r_{max}$ , where both $r_{min}$ and $r_{max}$ are known quantities. Given the aforementioned conditions, obviously $\beta=r_{min}$. However, I'm not quite sure how to go about solving for $\alpha$ or $s_i$. I was given the hint that the solution involves a Lambert $W$ function, but otherwise I'm in the dark. If anyone could potentially point me in the right direction, I would be eternally grateful!

Answer 1

We have $S_1=aS_0+b$ and $S_2=a^2S_0+ab$.

$S_{i+1}=aS_i+b~~~~(1)$, then $S_{i+2}=aS_{i+1}+b~~~~(2)$. Subtract (1) from (2} to get $$S_{i+2}-S_{i+1}=a (S_{i+1}-S_i)\implies T_{i+1}=aT_{i}\implies T_i=ca^i~~~(GP)$$ Further we have $$S_{i+1}-S_i=ca^i,~ c=(S_1-S_0)=(a-1)S_0+b$$ By telescoping, we get $$S_{k}-S_1=c[a+a^2+a^3+....a^{k-1}]=ac\frac{a^{k-1}-1}{a-1}$$ Finally, we get $$S_k=ac\frac{a^{k-1}-1}{a-1}+aS_0+b,~~~ c=(a-1)S_0+b$$