<s>[INST] what is the explicit form of this iterativ formular I am not sure, if there is an explicit form, but if there is, how do I get it?
This is the formula:
$$c_n=\frac{1-n \cdot c_{n-1}}{\lambda}$$
where $\lambda \in \mathbb{R}$ and $n \in \mathbb{N}$
I already tried some forms for c via trail and error, but I couldn't find a solution ...
 [/INST] We have $$\begin{align} c_{10} & = \tfrac 1\lambda - \tfrac{10}\lambda c_9 \\
& = \tfrac 1\lambda - \tfrac{10}{\lambda^2} (1-9c_8) \\
& = \tfrac 1\lambda - \tfrac{10}{\lambda^2} + \tfrac{10\cdot 9}{\lambda^2} c_8 \\
& = \tfrac 1\lambda - \tfrac{10}{\lambda^2} + \tfrac{10\cdot 9}{\lambda^3} -\tfrac{10\cdot 9\cdot 8}{\lambda^4}c_7 \\
& = \ldots = \tfrac 1\lambda - \tfrac{10}{\lambda^2} + \tfrac{10\cdot 9}{\lambda^3} -\tfrac{10\cdot 9\cdot 8}{\lambda^4}+\tfrac{10\cdot 9\cdot 8\cdot 7}{\lambda^5} - \ldots + (-1)^{10} \frac{10!}{\lambda^{11}}\cdot c_0 \end{align}$$
So if I have not done any mistakes, the solution is
$$c_n = \sum_{k=0}^{n-1} \left((-1)^k \frac{n!}{(n-k)!\cdot \lambda^{k+1}} \right) + (-1)^n \frac{n!}{\lambda^{n+1}} c_0$$
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