收集一个巧妙证法 有 f ( i ) = ∑ j = i n ( j i ) g ( j ) f(i)=\sum_{j=i}^n\binom{j}{i}g(j) f(i)=∑j=in(ij)g(j) 证 g ( i ) = ∑ j = i n ( − 1 ) j − i ( j i ) f ( i ) g(i)=\sum_{j=i}^n(-1)^{j-i}\binom{j}{i}f(i) g(i)=∑j=in(−1)j−i(ij)f(i) 考虑 f , g f,g f,g 的生成函数 G ( x ) = ∑ i = 0 n g ( i ) x i G(x)=\sum_{i=0}^ng(i)x^i G(x)=∑i=0ng(i)xi 则 G ( x + 1 ) = ∑ i = 0 n g ( i ) ∑ j = 0 i x j ( i j ) = ∑ j = 0 n x j ∑ i = j n ( i j ) g ( i ) = F ( x ) G(x+1)=\sum_{i=0}^ng(i)\sum_{j=0}^ix^j\binom{i}{j}=\sum_{j=0}^nx^j\sum_{i=j}^n\binom{i}{j}g(i)=F(x) G(x+1)=∑i=0ng(i)∑j=0ixj(ji)=∑j=0nxj∑i=jn(ji)g(i)=F(x) 所以 G ( x ) = F ( x − 1 ) = ∑ i = 0 n f ( i ) ∑ j = 0 i ( − 1 ) j − i ( i j ) x j = ∑ i = 0 n x i ∑ j = i n ( j i ) ( − 1 ) j − i f ( j ) G(x)=F(x-1)=\sum_{i=0}^nf(i)\sum_{j=0}^i(-1)^{j-i}\binom{i}{j}x^j=\sum_{i=0}^nx^i\sum_{j=i}^n\binom{j}{i}(-1)^{j-i}f(j) G(x)=F(x−1)=∑i=0nf(i)∑j=0i(−1)j−i(ji)xj=∑i=0nxi∑j=in(ij)(−1)j−if(j)