Let $Z_1,...,Z_S$ be the i.i.d. random variables with the following common cdf: $${F_Z}(x) = \sum\limits_{n = 0}^{N} {{b_n}{{\left( {1 - {e^{ - {c_n}x}}} \right)}^Q}} = \sum\limits_{n = 0}^N {\sum\limits_{q = 0}^Q {\left( \begin{array}{c} Q\\ q \end{array} \right)} {{( - 1)}^q}{b_n}{e^{ - q{c_n}x}}},$$ where $b_n>0$ and $c_n>0$ for $n =0,...,N$.
The cdf of the $k$th order statistic ($k$th largest among $Z_1,...,Z_S$) is given by \begin{align} {F_{{Z_{k:S}}}}(x) &= {\zeta _{k|S}}\sum\limits_{i = 0}^{S - k} {{S-k \choose i}\frac{{{{( - 1)}^i}}}{{k + i}}} {[{F_Z}(u)]^{k + i}}\\ &={\zeta _{k|S}} \sum\limits_{i = 0}^{S - k} {S-k \choose i}\frac{{{{( - 1)}^i}}}{{k + i}}\sum\limits_{\sum\limits_{v = 0}^{O-1} {{r_v}} = k + i} {k + i\choose {r_0}, ...,{r_{O-1}} } \\ &~~~\times \left(\prod\limits_{q = 0}^Q {{{\left( { {Q \choose q}{{\left( { - 1} \right)}^q}} \right)}^{\sum\limits_{n = 0}^N {{r_{n(Q + 1) + q}}} }}} \right)\left(\prod\limits_{n = 0}^N {b_n^{\sum\limits_{i = 0}^Q {{r_{n\left( {Q + 1} \right) + i}}} }} \right)\\&~~~\times {e^{ - x\sum\limits_{n = 0}^N {{c_n}\sum\limits_{i = 0}^Q {i{r_{n(Q + 1) + i}}} } }}, \end{align} where $\zeta_{k|S}= \frac{S!}{{(k - 1)!(S -k)!}}$ and $O = (N+1)(Q+1)$.
I tried to express this cdf as the following form: $$ {F_{{Z_{k:S}}}}(x) = \sum\limits_{n = 1}^{T} {{A}(n){e^{ - x{B}(n)}}}, $$ where $T$, $A(n)$, and $B(n)$ are needed to be specified. How do I specify these values and what is the general expression?
My approach: Without loss of generality, suppose that $B(1)< B(2)<....< B(T)$. Since $F_{Z}(\infty)=1$, $B(1)=0$ and $A(1)=1$. In order to specify $A(2)$, I use $$ \lim_{x\to\infty}{e^{xB(2)}}\left( {F(x) - A(1)} \right) = A(2) + \lim_{x\to\infty}\sum\limits_{n = 2}^T {A(n){e^{ - x\left( {B(n) - B(2)} \right)}}} =A(2), $$ where the last equality comes from the fact $B(2)<B(n)$ for $n>2$. Recursively, I can specify $A(3)$, $A(4)$,..., and $A(T)$. This approach requires the sorting operation on $\{B(n)\}$ so that it is hard to obtain the general expression of $A(n)$.