徐利治教授在《数学分析的方法及例题选讲》一书中,给出了一个组合恒等式,即 multiply from k=1 to n(-1)~(n-k)C_n~kk~n=n!。(1)这是有名的Euler恒等式。本文将恒等式(1)作了如下推广: multiply from k=0 to n(-1)~(n-k)C_n~ka_(k-1)~n=n!d~n。(2)其中a_1,a_2,…,a_(n+1)是以d为公差的等差数列。显然,在(2)中,当d=1,a_1=0时,便可得到恒等式(1)。为了证明(2),需要下面的结论。引理下述恒等式成立。 multiply from k=0 to n(-1)~(n-k)C_n~kk~m=0 (3)其中n>m≥0,并约定0~0=1。证明因为 multiply from k=0 to n(-1)~(n-k)C_n~kk~m =multiply from k=0 to n(-1)~(n-k)(-1)~(2k)C_n~kk~m Prof. Xu Lizhi gave a combinatorial equivalence in the book “Methods of Mathematical Analysis and Selected Examples”, that is, multiply from k=1 to n(-1)~(n-k)C_n~kk~n=n!. (1) This is the famous Euler identity. In this paper, the identities (1) are generalized as follows: multiply from k=0 to n(-1)~(n-k)C_n~ka_(k-1)~n=n!d~n. (2) where a_1, a_2, ..., a_(n+1) are arithmetic progressions with d as a tolerance. Obviously, in (2), when d=1 and a_1=0, identity (1) can be obtained. To prove (2), the following conclusions are needed. The lemma of the following identities holds. Multiply from k=0 to n(-1)~(n-k)C_n~kk~m=0 (3) where n>m≥0 and agree on 0~0=1. Prove that multiply from k=0 to n(-1)~(nk)C_n~kk~m =multiply from k=0 to n(-1)~(nk)(-1)~(2k)C_n~kk~ m