# 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?

The source of the expression is probabilistic calculation.

• Well, it's equal to $$\sum_{n=1}^x\sum_{m=n}^x \frac{(2a)^{x-m}}{(x-m)!}$$ – Don Thousand Dec 17 '18 at 17:09
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 $$Q(s,z) = {{\Gamma (s,z)} \over {\Gamma (s)}} = {{\sum\limits_{k = 0}^{s - 1} {{{z^{\,k} } \over {k!}}} } \over {e^{\,z} }}$$