Expansion coefficients Suppose that we are given the function $f(x)$ in the following product form: 
$$f(x) = \prod_{k = -K}^K (1-a^k x)\,,$$
Where $a$ is some real number. 
I would like to find the expansion coefficients $c_n$, such that:
$$f(x) = \sum_{n = 0}^{2K+1} c_n x^n\,.$$
A closed form solution for $c_n$, or at least a relation between the coefficients $c_n$ (e.g. between $c_n$ and $c_{n+1}$) would be great! 

We have that
$$ \bbox[lightyellow] {  
\eqalign{
  & f(x) = \prod\limits_{k\, = \, - K}^K {\left( {1 - a^{\;k} x} \right)}  = \prod\limits_{k\, = \, - K}^K {a^{\;k} \left( {a^{\; - k}  - x} \right)}  =   \cr 
  &  = \left( {\prod\limits_{k\, = \, - K}^K {a^{\;k} } } \right)\;\prod\limits_{k\, = \, - K}^K {\left( {a^{\; - k}  - x} \right)}  = \left( {\prod\limits_{k\, = \, - K}^K {a^{\;k} } } \right)\;\left( { - 1} \right)^{2K + 1} \prod\limits_{k\, = \, - K}^K {\left( {x - a^{\; - k} } \right)}  =   \cr 
  &  =  - \left( {\prod\limits_{k\, = \, - K}^K {a^{\;k} } } \right)\;\prod\limits_{k\, = \, - K}^K {\left( {x - a^{\; - k} } \right)}  =  - \prod\limits_{k\, = \, - K}^K {\left( {x - a^{\; - k} } \right)}  =  - \prod\limits_{k\, = \, - K}^K {\left( {x - a^{\;k} } \right)}  =   \cr 
  &  =  - \left( {x - 1} \right)\prod\limits_{k\, = \,1}^K {\left( {x - a^{\;k} } \right)} \prod\limits_{k\, = \,1}^K {\left( {x - \left( {{1 \over a}} \right)^{\;k} } \right)}  =   \cr 
  &  =  - {1 \over {\left( {x - 1} \right)}}\prod\limits_{k\, = \,0}^K {\left( {x - a^{\;k} } \right)} \prod\limits_{k\, = \,0}^K {\left( {x - \left( {{1 \over a}} \right)^{\;k} } \right)}  =   \cr 
  &  = {{\left( { - 1} \right)^{\,K} } \over {\left( {x - 1} \right)}}a^{\, - \,\left( {\scriptstyle K + 1 \atop 
  \scriptstyle 2}  \right)} \left( {\prod\limits_{k\, = \,0}^K {\left( {x - a^{\;k} } \right)} } \right)^{\,2}  \cr} 
 }$$
because
$$
\prod\limits_{k\, = \,0}^K {\left( {x - a^{\;k} } \right)}  = \left( { - 1} \right)^{\,K + 1} a^{\,\left( {\scriptstyle K + 1 \atop 
  \scriptstyle 2}  \right)} \prod\limits_{k\, = \,0}^K {\left( {x - \left( {{1 \over a}} \right)^{\;k} } \right)} 
$$
Then the
$$ \bbox[lightyellow] {  
\prod\limits_{k\, = \,0}^{n - 1} {\left( {1 - xq^{\;k} } \right)} 
 }$$
is the q-Pochammer Symbol
$(x;\,q)_n$.
We can get the coefficients of the polynomial
$$ \bbox[lightyellow] {  
f(x) =  - \prod\limits_{k\, = \, - K}^K {\left( {x - a^{\;k} } \right)}  =  - \sum\limits_{j\, = \,0}^{2K + 1} {c_{\,j} \;x^{\,j} } 
 }$$
by means of the Vieta's Formulas
$$ \bbox[lightyellow] {  
\left\{ \matrix{
  c_{\;2K + 1}  = 1 \hfill \cr 
  \left( { - 1} \right)^{\,k} c_{\;2K + 1 - k} \quad \left| {\;1 \le k} \right.\quad  = \sum\limits_{ - K\, \le \,j_{\,1} \, < \,j_{\,2} \, < \, \cdots \, < \,j_{\,k} \, \le \,K} {a^{\;j_{\,1} \, + \,j_{\,2} \, + \, \cdots \, + \,j_{\,k} } }  \hfill \cr}  \right.
 }$$
Now
$$ \bbox[lightyellow] {  
\sum\limits_{\left\{ \begin{subarray}{l} 
  \;1\, \leqslant \,l\, \leqslant \,k \\ 
  \; - K\, \leqslant \,j_{\,1} \, < \,j_{\,2} \, < \, \cdots \, < \,j_{\,k} \, \leqslant \,K 
\end{subarray}  \right.} {\;j_{\,l} \,\,}  =  - k\left( {K + 1} \right) + \sum\limits_{\left\{ \begin{subarray}{l} 
  \;1\, \leqslant \,l\, \leqslant \,k \\ 
  \;1\, \leqslant \,j_{\,1} \, < \,j_{\,2} \, < \, \cdots \, < \,j_{\,k} \, \leqslant \,2K + 1 
\end{subarray}  \right.} {\;j_{\,l} \,\,} 
 }$$
and we are left to evaluate the last sum.
That is, the sum of the elements of all the k-subsets of the set $\left\{ {1,\,2,\, \cdots ,\,2K + 1} \right\}$.
The total number of k-subsets is
$$
N_{k - subs}  = \left( \matrix{
  2K + 1 \cr 
  k \cr}  \right)
$$
The number of k-subsets containing a given element $j$ clearly is
$$
N_{j\, \in \;k - subs}  = \left( \matrix{
  2K \cr 
  k - 1 \cr}  \right)
$$
Thus the sum we are looking for is
$$ \bbox[lightyellow] {  
\begin{gathered}
  S(k,2K + 1) =  - k\left( {K + 1} \right) + \sum\limits_{\left\{ \begin{subarray}{l} 
  \;1\, \leqslant \,l\, \leqslant \,k \\ 
  \;1\, \leqslant \,j_{\,1} \, < \,j_{\,2} \, < \, \cdots \, < \,j_{\,k} \, \leqslant \,2K + 1 
\end{subarray}  \right.} {\;j_{\,l} \,\,}  =  \hfill \\
   =  - k\left( {K + 1} \right) + \left( \begin{gathered}
  2K \\ 
  k - 1 \\ 
\end{gathered}  \right)\sum\limits_{\;1\, \leqslant \,j\, \leqslant \,2K + 1} {\;j\,}  = \left( \begin{gathered}
  2K \\ 
  k - 1 \\ 
\end{gathered}  \right)\left( \begin{gathered}
  2K + 2 \\ 
  2 \\ 
\end{gathered}  \right) - k\left( {K + 1} \right) \hfill \\ 
\end{gathered} 
 }$$
