这是一类有关数列求和的问题,核心的解题思路在于:kcnk = ncn-1k-1.下面是应用的例子: 求:∑k=0n(x+k)2cnkxk(1-x)n-k 的值. 解:∑k=0n(x+k)2cnkxk(1-x)n-k = ∑k=0n(x2+2kx+k2) cnkxk(1-x)n-k = ∑k=0nx2cnkxk(1-x)n-k + ∑k=0n2kx cnkxk(1-x)n-k + ∑k=0nk2 cnkxk(1-x)n-k. 分开求解: ∑k=0nx2cnkxk(1-x)n-k = x2∑k=1ncnkxk(1-x)n-k = x2(x+1-x)n = x2 ; kcnk = ncn-1k-1 ∑k=0n2kx cnkxk(1-x)n-k = 2nx2∑k=1ncn-1k-1xk-1(1-x)n-k = 2nx2(x+1-x)n-1 = 2nx2; k(k-1)cnk = n(n-1)cn-2k-2;