m***x
发帖数: 4 | 1 \sum_{k=0} ^{n} (-1)^k C(n,k) f(k)
is to take the n-th divided difference of f(k)
denote it by DD^n f(k) where
DD f(k)= f(k)-f(k-1).
DD takes a polynomial in k of degree m in to a polynomial in k
of degree m-1. So what you want is a constant.
now you need only to prove
DD^n k(k+1)...(k+n-1)=n!,
which is easily done by induction. |
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