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Is the value of the following summation can be expressed in a closed form?

$$ \Delta = \sum_{k=0}^{n} S(n,k) \; (C)_k \; m^k $$

where $S(n,k)$ is Stirling number of second kind, $(C)_k$ is the falling factorial defined as $(C)_k:=C(C-1)\cdots (C-k+1)$ and $ 0< m <1$.

farzada
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  • I just noticed that we can redefine the falling factorial as $(C)k = {C \choose k} n!$ and arrive at $$ \Delta = \sum{k=1}^n k!S(n, k){C \choose k} m^k. $$ where it is n-th moment of the binomial random variable posted at http://math.stackexchange.com/questions/1488422/a-sum-involving-stirling-numbers-of-the-second-kind?rq=1. Is this possible to get a close expression for it? – farzada Dec 17 '16 at 12:42
  • A generalization of this sum is posted at http://math.stackexchange.com/questions/1486986/a-combinatorial-sum-and-identity-involving-stirling-numbers-of-the-second-kind – farzada Dec 17 '16 at 13:53
  • This question is duplicate of http://math.stackexchange.com/questions/1735707/is-there-a-closed-form-expression-for-this-sum-involving-stirling-number-of-seco – farzada Dec 17 '16 at 13:56

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