Given the following: $$\begin{align} v_n&=v_{n-1}-r\sum_{i=n}^m \frac p{v_i}\\ v_0&=V \end{align}$$ where: $$\begin{align} m\ge n\ge 0\;&:\;m,\,n\in\mathbb {N_0}\\ V,\,r,\,p,\,m\;&:\;\text{constants} \end{align}$$ I'm trying to determine an expression for $v_n$ that does not involve $v_n$ on the right-hand side. (As presented above, the definition is circular.)
Mathematica's
Solve function can solve this for a specific value of $m$ if I write out the equations. For example:
$$\begin{align}
v_0&=V\\
v_1&=v_0-r\,(\frac p{v_1}+\frac p{v_2}+\frac p{v_3})\\
v_2&=v_1-r\,(\frac p{v_2}+\frac p{v_3})\\
v_3&=v_2-r\,\frac p{v_3}
\end{align}$$
Solved symbolically, the result is pages and pages long and involves many Root objects. Numerically, with $V=48$, $p=36$, and $r=3$, the result is:
$$\begin{align}
v_0&=48.0000\\
v_1&=36.7473\\
v_2&=28.4335\\
v_3&=23.9181
\end{align}$$
If possible, I would like to derive a more general solution (for arbitrary $m$) that is compact enough to represent reasonably in Excel (or maybe C). It doesn't have to be a closed-form expression (though that is prefered); it could be a simple algorithm.
In trying to solve this, I've discovered recurrence relations (and Mathematica's RSolve function), which this looks a lot like to me, but I'm not versed in those, and I have no idea how to apply those to solving this problem (or whether that is even possible). (Based on what I've seen so far, recurrence relations seem to deal mainly with infinite sequences, whereas this is very much a finite, but arbitrary-length, sequence, and each node depends on the values of the complete sequence.)
Background
This is an electronics problem.
I have a number of constant-power loads at regular intervals along a cable, and I want to determine the voltage at each load, taking into account the resistive voltage drop of the cable.
Here's a schematic representation:
Because the loads are constant-power, the current they draw increases as their voltage decreases. The increased current, in turn, decreases the voltage (through the resistive cable losses), which further increases the current.