∑j=kn∑i′=0j−k(−1)j−i′−k(jk)(j−ki′)f(j)=∑j=kn(1−1)j−k(jk)f(j)(−1)j−k\sum^n_{j=k}\sum^{j-k}_{i'=0}(-1)^{j-i'-k}\tbinom{j}{k}\tbinom{j-k}{i'}f(j)=\sum_{j=k}^n(1-1)^{j-k}\tbinom{j}{k}f(j)(-1)^{j-k}∑j=kn∑i′=0j−k(−1)j−i′−k(kj)(i′j−k)f(j)=∑j=kn(1−1)j−k(kj)f(j)(−1)j−k 怎么用二项式定理推
