Is there a way to simplify the sum \begin{align*} \sum_{m=0}^k \binom{k}{m} \underbrace{q(q-1)\dots(q-k+m+1)}_{k-m\text{ terms}}(-1)^{k-m} (q-1)^m, \end{align*}

where $q$ is sufficiently large? It clearly resembles the binomial expansion, as well as this falling factorial version of it, but it is different from both.

If there is no easy way to simplify it, a good upper bound, or a characterization of scaling behavior with $q$ would also be useful.


1 Answer 1


I am not sure that this would help much.

Let $$a_m=(-1)^{k-m}\left(\prod_{i=0}^{k-m-1}(q-i)\right)\binom{k}{m}(q-1)^m=-q (q-1)^m \binom{k}{m} (1-q)_{k-m-1}$$ where appears Pochhammer symbol $$\sum_{m=0}^ka_m=\frac{ \Gamma (k-q)}{\Gamma (-q)}\, _1F_1(-k;-k+q+1;q-1)$$where appears Kummer confluent hypergeometric function.


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