Can $\sum_{n=1}^{x}n\frac{(2a)^{x-n}}{(x-n)!}$ be simplified? I was wondering if it is possible to simplify the following expression:
$$\sum_{n=1}^{x}n\frac{(2a)^{x-n}}{(x-n)!}$$
Where $0<a<1$?
The source of the expression is probabilistic calculation.
Can anyone please help?
 A: You can write it as
$$
\eqalign{
  & \sum\limits_{1\, \le \,n\, \le \,x} {n{{\left( {2a} \right)^{\,x - n} } \over {\left( {x - n} \right)!}}}
  = \sum\limits_{0\, \le \,n\, \le \,x} {n{{\left( {2a} \right)^{\,x - n} } \over {\left( {x - n} \right)!}}}  =   \cr 
  &  = \sum\limits_{0\, \le \,n\, \le \,x} {\left( {x - \left( {x - n} \right)} \right){{\left( {2a} \right)^{\,x - n} } \over {\left( {x - n} \right)!}}}  =   \cr 
  &  = \sum\limits_{0\, \le \,k\, \le \,x} {\left( {x - k} \right){{\left( {2a} \right)^{\,k} } \over {k!}}}  =   \cr 
  &  = x\sum\limits_{0\, \le \,k\, \le \,x} {{{\left( {2a} \right)^{\,k} } \over {k!}}}  - \sum\limits_{1\, \le \,k\, \le \,x} {k{{\left( {2a} \right)^{\,k} } \over {k!}}}  =   \cr 
  &  = x\sum\limits_{0\, \le \,k\, \le \,x} {{{\left( {2a} \right)^{\,k} } \over {k!}}}  - 2a\sum\limits_{1\, \le \,k\, \le \,x} {{{\left( {2a} \right)^{\,k - 1} }
  \over {\left( {k - 1} \right)!}}}  =   \cr 
  &  = x\sum\limits_{0\, \le \,k\, \le \,x} {{{\left( {2a} \right)^{\,k} } \over {k!}}}  - 2a\sum\limits_{0\, \le \,k\, \le \,x - 1} {{{\left( {2a} \right)^{\,k} } \over {k!}}}  =   \cr 
  &  = x\left( {\sum\limits_{0\, \le \,k\, \le \,x - 1} {{{\left( {2a} \right)^{\,k} } \over {k!}}}
  + {{\left( {2a} \right)^{\,x} } \over {x!}}} \right) - 2a\sum\limits_{0\, \le \,k\, \le \,x - 1} {{{\left( {2a} \right)^{\,k} } \over {k!}}}  =   \cr 
  &  = x{{\left( {2a} \right)^{\,x} } \over {x!}} + \left( {x - 2a} \right)\sum\limits_{0\, \le \,k\, \le \,x - 1} {{{\left( {2a} \right)^{\,k} } \over {k!}}}  =   \cr 
 &  = {{\left( {2a} \right)^{\,x} } \over {\Gamma (x)}} + \left( {x - 2a} \right)e^{\,2a} Q(x,2a)\quad \left| {\;a,x \in \mathbb C} \right. \cr} 
$$
where $Q(s,z)$ is the Regularized Incomplete Upper Gamma
$$
Q(s,z) = {{\Gamma (s,z)} \over {\Gamma (s)}} = {{\sum\limits_{k = 0}^{s - 1} {{{z^{\,k} } \over {k!}}} } \over {e^{\,z} }}
$$
Also refer to this WF article and to this related post.
