I was wondering whether the following relationship could be solved.
$$S_0= 100, C=(0.05, 0.15,...), P = (0.03, 0.05, ...), D = 120, R = 0.08\\ S_1 = S_0(1-C_1)\left(1-P_1-\frac{1}{D + 1}-(1-R)\right)\\ S_2 = S_1(1-C_2)\left(1-P_2-\frac{1}{D + 1}-(1-R)\right)\\ S_3 = ...$$
Notice that the $C_1$ refers to the first element in the $C$ vector and $P_1$ refers to the first element in the $P$ vector, while $D$ and $R$ are constants.
I have tried to come up with a part, but I can't seem to solve the latter part, which is in the latter brackets.
$$S_n=S_0\cdot\Pi_{i=1}^{n}(1-C_i)\cdot ...$$
But the latter part, I am not sure how to do it, except just writing it out, but then I am still stuck unfortunately. I have the feeling that it can be solved, but I am not sure how.
Please do not solve the $C$ and $P$ they can be random and are considered user input, so they can vary arbitrary.
I think it might be possible to solve this recurrence relation. The first thing to do is find out an expression for $C_n$ and one for $P_n.$ I get \begin{align*} C_{n+1}&=0.05+0.1n \\ P_{n+1}&=0.03+0.02n. \end{align*} The recurrence relation we can write as \begin{align*} S_{n+1} &= S_n(1-C_{n+1})\left(1-P_{n+1}-\frac{1}{D + 1}-(1-R)\right) \\ &=S_n(1-(0.05+0.1n))\left(1-(0.03+0.02n)-\frac{1}{121}-0.92\right) \\ &=S_n(0.95-0.1n)\left(0.05-0.02n-\frac{1}{121}\right). \end{align*} This is a difficult recurrence relation to solve, but Mathematica comes through with the following: $$S_n=\frac{1919}{121}\, 4^{-n} 125^{1-n} \left(-\frac{17}{2}\right)_{n-1} \left(-\frac{263}{242}\right)_{n-1}, $$ where the symbol $\displaystyle\left(-\frac{17}{2}\right)_{n-1} $ is the Pochhammer symbol.