组合数恒等式

hbghlyj

$p\in\mathbf N_+,c\in\mathbf R,$求证：
\[\sum\limits_{k=0}^p (-1)^k \left(\binom{p+1}{k} (c-k+p+1)^{p+1}-(p+1) \binom{p}{k} (c-k+p)^p\right)=-(-c)^{p+1}\]

tommywong

$\displaystyle \sum_{k=0}^p (-1)^k\binom{p+1}{k}(c-k+p+1)^{p+1}$
$\displaystyle =-(-c)^{p+1}
+\sum_{k=0}^{p+1} (-1)^{p+1-k}\binom{p+1}{k}(c+k)^{p+1}$
$=-(-c)^{p+1}+\Delta^{p+1} (c+k)^{p+1}=-(-c)^{p+1}+(p+1)!$

$\displaystyle \sum_{k=0}^p (-1)^k\binom{p}{k}(c-k+p)^p
=\Delta^p (c+k)^p=p!$