What is the general formula of the sum $\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{k/2}{m}$ for $m,n\in\mathbb{N}$?

Solution 1:

Let $\mathbb{N}_0=\{0,1,2,\dotsc\}$.

• For $n,\ell\in\mathbb{N}_0$, we have \begin{equation}\tag{1} \sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{k/2}{\ell} = \begin{cases} 0, & n>\ell\in\mathbb{N}_0;\\ \displaystyle (-1)^{\ell}n!\frac{[2(\ell-n)-1]!!}{(2\ell)!!}\binom{2\ell-n-1}{2(\ell-n)}, & \ell\ge n\in\mathbb{N}_0. \end{cases} \end{equation}
• For $n,\ell\in\mathbb{N}_0$, we have \begin{equation}\tag{2} \sum_{k=0}^{n}(-1)^{k}\binom{n}{k}\binom{2k}{\ell} = \begin{cases} 0, & n>\ell\in\mathbb{N}_0;\\\displaystyle (-1)^n\binom{n}{\ell-n}2^{2n-\ell}, & \ell\ge n\in\mathbb{N}_0. \end{cases} \end{equation}
• For $\ell\ge n\in\mathbb{N}_0$, we have \begin{equation}\tag{3} \sum_{n=0}^\ell\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{k/2}{\ell} =(-1)^\ell\frac{(2\ell-1)!!}{(2\ell)!!}. \end{equation}

Reference

1. Siqintuya Jin, Bai-Ni Guo, and Feng Qi, Partial Bell polynomials, falling and rising factorials, Stirling numbers, and combinatorial identities, Computer Modeling in Engineering & Sciences (2022), in press; accepted on 24 January 2022; available online at https://dx.doi.org/10.32604/cmes.2022.019941 or https://www.researchgate.net/publication/358050501.