计算机程序设计艺术习题解答（Excercise 1.2.3-30～34题）

30. [M23 ] (J. Binet, 1812.) 不使用归纳法，证明等式 

证：

这个公式是一个非常重要且基础的通用公式，稍后可以看到它的一系列推论，用处很大。

首先根据正文里公式 (4)，用于和式乘积的分配律 (∑R(i)ai)(∑S(j)bj)=∑R(i)∑S(j)aibj" role="presentation" style="position: relative;">(∑R(i)ai)(∑S(j)bj)=∑R(i)∑S(j)aibj$(\sum _{R(i)}{a}_i)(\sum _{S(j)}{b}_j)=\sum _{R(i)}\sum _{S(j)}{a}_i{b}_j$ ，我们有

(∑j=1najxj)(∑j=1nbjyj)=∑j=1n∑k=1najxjbkyk=∑1≤j≤n∑1≤k≤najxjbkyk=∑1≤j,k≤najbkxjyk" role="presentation" style="position: relative;">(∑nj=1ajxj)(∑nj=1bjyj)=∑nj=1∑nk=1ajxjbkyk=∑1≤j≤n∑1≤k≤najxjbkyk=∑1≤j,k≤najbkxjyk$(\sum _{j=1}^n{a}_j{x}_j)(\sum _{j=1}^n{b}_j{y}_j)=\sum _{j=1}^n\sum _{k=1}^n{a}_j{x}_j{b}_k{y}_k=\sum _{1\le j\le n}\sum _{1\le k\le n}{a}_j{x}_j{b}_k{y}_k=\sum _{1\le j,k\le n}{a}_j{b}_k{x}_j{y}_k$  ，

同理有 (∑j=1najyj)(∑j=1nbjxj)=∑1≤j,k≤najbkxkyj" role="presentation" style="position: relative;">(∑nj=1ajyj)(∑nj=1bjxj)=∑1≤j,k≤najbkxkyj$(\sum _{j=1}^n{a}_j{y}_j)(\sum _{j=1}^n{b}_j{x}_j)=\sum _{1\le j,k\le n}{a}_j{b}_k{x}_k{y}_j$

上下两个等式相减，得到

(∑j=1najxj)(∑j=1nbjyj)−(∑j=1najyj)(∑j=1nbjxj)=∑1≤j,k≤n(ajbkxjyk−ajbkxkyj)" role="presentation" style="position: relative;">(∑nj=1ajxj)(∑nj=1bjyj)−(∑nj=1ajyj)(∑nj=1bjxj)=∑1≤j,k≤n(ajbkxjyk−ajbkxkyj)$(\sum _{j=1}^n{a}_j{x}_j)(\sum _{j=1}^n{b}_j{y}_j)-(\sum _{j=1}^n{a}_j{y}_j)(\sum _{j=1}^n{b}_j{x}_j)=\sum _{1\le j,k\le n}(a_j{b}_k{x}_j{y}_k-a_j{b}_k{x}_k{y}_j)$

= ∑1≤j<k≤n(ajbkxjyk−ajbkxkyj)+∑1≤k<j≤n(ajbkxjyk−ajbkxkyj)" role="presentation" style="position: relative;">∑1≤j<k≤n(ajbkxjyk−ajbkxkyj)+∑1≤k<j≤n(ajbkxjyk−ajbkxkyj)$\sum _{1\le j<k\le n}(a_j{b}_k{x}_j{y}_k-a_j{b}_k{x}_k{y}_j)+\sum _{1\le k<j\le n}(a_j{b}_k{x}_j{y}_k-a_j{b}_k{x}_k{y}_j)$  ( 1≤j=k≤n" role="presentation" style="position: relative;">1≤j=k≤n$1\le j=k\le n$ 时和式为零)

= ∑1≤j<k≤n(ajbkxjyk−ajbkxkyj)+∑1≤j<k≤n(akbjxkyj−ajbkxkyj)" role="presentation" style="position: relative;">∑1≤j<k≤n(ajbkxjyk−ajbkxkyj)+∑1≤j<k≤n(akbjxkyj−ajbkxkyj)$\sum _{1\le j<k\le n}(a_j{b}_k{x}_j{y}_k-a_j{b}_k{x}_k{y}_j)+\sum _{1\le j<k\le n}(a_k{b}_j{x}_k{y}_j-a_j{b}_k{x}_k{y}_j)$ （第二个和式中 j 与 k 互换）

= ∑1≤j<k≤n(ajbk(xjyk−xkyj))−∑1≤j<k≤n(akbj(xjyk−xkyj))" role="presentation" style="position: relative;">∑1≤j<k≤n(ajbk(xjyk−xkyj))−∑1≤j<k≤n(akbj(xjyk−xkyj))$\sum _{1\le j<k\le n}(a_j{b}_k(x_j{y}_k-x_k{y}_j))-\sum _{1\le j<k\le n}(a_k{b}_j(x_j{y}_k-x_k{y}_j))$

= ∑1≤j<k≤n(ajbk−akbj)(xjyk−xkyj)" role="presentation" style="position: relative;">∑1≤j<k≤n(ajbk−akbj)(xjyk−xkyj)$\sum _{1\le j<k\le n}(a_j{b}_k-a_k{b}_j)(x_j{y}_k-x_k{y}_j)$       由此等式  得证。

一个重要的特例是：当 ω1,...,ωn,z1,...,zn" role="presentation" style="position: relative;">ω1,...,ωn,z1,...,zn${\omega }_1,...,{\omega }_n,z_1,...,z_n$ 为任意复数时，我们令 aj=ωj,bj=zj¯,xj=ωj¯,yj=zj" role="presentation" style="position: relative;">aj=ωj,bj=zj¯,xj=ωj¯,yj=zj$a_j={\omega }_j,b_j=\overline{z_j},x_j=\overline{{\omega }_j},y_j=z_j$ ，就得到等式 (∑j=1n|ωj|2)(∑j=1n|zj|2)=|∑j=1nωjzj|2+∑1≤j<k≤n|ωjzk¯−ωkzj¯|2" role="presentation" style="position: relative;">(∑nj=1|ωj|2)(∑nj=1|zj|2)=∣∣∑nj=1ωjzj∣∣2+∑1≤j<k≤n|ωjzk¯−ωkzj¯|2$(\sum _{j=1}^n{\left|{\omega }_j\right|}^2)(\sum _{j=1}^n{\left|z_j\right|}^2)={\left|\sum _{j=1}^n{\omega }_j{z}_j\right|}^2+\sum _{1\le j<k\le n}{\left|{\omega }_j\overline{z_k}-{\omega }_k\overline{z_j}\right|}^2$

证明：等式左边   (∑j=1najxj)(∑j=1nbjyj)=(∑j=1nωjωj¯)(∑j=1nzjzj¯)=(∑j=1n|ωj|2)(∑j=1n|zj|2)" role="presentation" style="position: relative;">(∑nj=1ajxj)(∑nj=1bjyj)=(∑nj=1ωjωj¯)(∑nj=1zjzj¯)=(∑nj=1|ωj|2)(∑nj=1|zj|2)$(\sum _{j=1}^n{a}_j{x}_j)(\sum _{j=1}^n{b}_j{y}_j)=(\sum _{j=1}^n{\omega }_j\overline{{\omega }_j})(\sum _{j=1}^n{z}_j\overline{z_j})=(\sum _{j=1}^n{\left|{\omega }_j\right|}^2)(\sum _{j=1}^n{\left|z_j\right|}^2)$

等式右边  (∑j=1najyj)(∑j=1nbjxj)+∑1≤j<k≤n(ajbk−akbj)(xjyk−xkyj)" role="presentation" style="position: relative;">(∑nj=1ajyj)(∑nj=1bjxj)+∑1≤j<k≤n(ajbk−akbj)(xjyk−xkyj)$(\sum _{j=1}^n{a}_j{y}_j)(\sum _{j=1}^n{b}_j{x}_j)+\sum _{1\le j<k\le n}(a_j{b}_k-a_k{b}_j)(x_j{y}_k-x_k{y}_j)$

=  (∑j=1nωjzj)(∑j=1nωj¯zj¯)+∑1≤j<k≤n(ωjzk¯−ωkzj¯)(ωj¯zk−ωk¯zj)" role="presentation" style="position: relative;">(∑nj=1ωjzj)(∑nj=1ωj¯zj¯)+∑1≤j<k≤n(ωjzk¯−ωkzj¯)(ωj¯zk−ωk¯zj)$(\sum _{j=1}^n{\omega }_j{z}_j)(\sum _{j=1}^n\overline{{\omega }_j}\overline{z_j})+\sum _{1\le j<k\le n}({\omega }_j\overline{z_k}-{\omega }_k\overline{z_j})(\overline{{\omega }_j}{z}_k-\overline{{\omega }_k}{z}_j)$

=  |∑j=1nωjzj|2+∑1≤j<k≤n|ωjzk¯−ωkzj¯|2" role="presentation" style="position: relative;">∣∣∑nj=1ωjzj∣∣2+∑1≤j<k≤n|ωjzk¯−ωkzj¯|2${\left|\sum _{j=1}^n{\omega }_j{z}_j\right|}^2+\sum _{1\le j<k\le n}{\left|{\omega }_j\overline{z_k}-{\omega }_k\overline{z_j}\right|}^2$              

 注：    (∑j=1nωjzj)" role="presentation" style="position: relative;">(∑nj=1ωjzj)$(\sum _{j=1}^n{\omega }_j{z}_j)$ 与 (∑j=1nωj¯zj¯)" role="presentation" style="position: relative;">(∑nj=1ωj¯zj¯)$(\sum _{j=1}^n\overline{{\omega }_j}\overline{z_j})$ 互为共轭，(ωjzk¯−ωkzj¯)" role="presentation" style="position: relative;">(ωjzk¯−ωkzj¯)$({\omega }_j\overline{z_k}-{\omega }_k\overline{z_j})$ 与 (ωj¯zk−ωk¯zj)" role="presentation" style="position: relative;">(ωj¯zk−ωk¯zj)$(\overline{{\omega }_j}{z}_k-\overline{{\omega }_k}{z}_j)$ 也是互为共轭关系。

所有的 (|ωjzk¯−ωkzj¯|2)" role="presentation" style="position: relative;">(|ωjzk¯−ωkzj¯|2)$({\left|{\omega }_j\overline{z_k}-{\omega }_k\overline{z_j}\right|}^2)$ 项都是非负的，所以根据 Binet 公式就可以推出著名的柯西-施瓦茨(Cauchy-Schwarz) 不等式 (∑j=1n|ωj|2)(∑j=1n|zj|2)≥|∑j=1nωjzj|2" role="presentation" style="position: relative;">(∑nj=1|ωj|2)(∑nj=1|zj|2)≥∣∣∑nj=1ωjzj∣∣2$(\sum _{j=1}^n{\left|{\omega }_j\right|}^2)(\sum _{j=1}^n{\left|z_j\right|}^2)\ge {\left|\sum _{j=1}^n{\omega }_j{z}_j\right|}^2$ 。

31. 【M20] 利用 Binet 公式将和式 ∑1≤j<k≤n(uj−uk)(vj−vk)" role="presentation" style="position: relative;">∑1≤j<k≤n(uj−uk)(vj−vk)$\sum _{1\le j<k\le n}(u_j-u_k)(v_j-v_k)$ 表示为由 ∑j=1nujvj,∑j=1nujand∑j=1nvj" role="presentation" style="position: relative;">∑nj=1ujvj,∑nj=1ujand∑nj=1vj$\sum _{j=1}^n{u}_j{v}_j,\sum _{j=1}^n{u}_{j}and\sum _{j=1}^n{v}_j$ 组成的式子。

解：

在 Binet 公式  中，令 aj=uj,bk=1,xj=vj,yk=1" role="presentation" style="position: relative;">aj=uj,bk=1,xj=vj,yk=1$a_j=u_j,b_k=1,x_j=v_j,y_k=1$ ，就得到 ∑1≤j<k≤n(uj−uk)(vj−vk)=(∑j=1nujvj)(∑j=1n1⋅1)−(∑j=1nuj⋅1)(∑j=1nvj⋅1)" role="presentation" style="position: relative;">∑1≤j<k≤n(uj−uk)(vj−vk)=(∑nj=1ujvj)(∑nj=11⋅1)−(∑nj=1uj⋅1)(∑nj=1vj⋅1)$\sum _{1\le j<k\le n}(u_j-u_k)(v_j-v_k)=(\sum _{j=1}^n{u}_j{v}_j)(\sum _{j=1}^{n}1\cdot 1)-(\sum _{j=1}^n{u}_j\cdot 1)(\sum _{j=1}^n{v}_j\cdot 1)$

= n(∑j=1nujvj)−(∑j=1nuj)(∑j=1nvj)" role="presentation" style="position: relative;">n(∑nj=1ujvj)−(∑nj=1uj)(∑nj=1vj)$n(\sum _{j=1}^n{u}_j{v}_j)-(\sum _{j=1}^n{u}_j)(\sum _{j=1}^n{v}_j)$

当 uj" role="presentation" style="position: relative;">uj$u_j$ 和 vj" role="presentation" style="position: relative;">vj$v_j$ 同为单调递增数列时，即 u1≤u2≤...≤un" role="presentation" style="position: relative;">u1≤u2≤...≤un$u_1\le u_2\le ...\le u_n$ 且 v1≤v2≤...≤vn" role="presentation" style="position: relative;">v1≤v2≤...≤vn$v_1\le v_2\le ...\le v_n$ 时，对任何 j, k, 只要满足 1≤j<k≤n" role="presentation" style="position: relative;">1≤j<k≤n$1\le j<k\le n$ ，都有 (uj−uk)(vj−vk)≥0" role="presentation" style="position: relative;">(uj−uk)(vj−vk)≥0$(u_j-u_k)(v_j-v_k)\ge 0$ ，这样就可以推导出 ∑1≤j<k≤n(uj−uk)(vj−vk)≥0⇒(∑j=1nuj)(∑j=1nvj)≤n(∑j=1nujvj)" role="presentation" style="position: relative;">∑1≤j<k≤n(uj−uk)(vj−vk)≥0⇒(∑nj=1uj)(∑nj=1vj)≤n(∑nj=1ujvj)$\sum _{1\le j<k\le n}(u_j-u_k)(v_j-v_k)\ge 0\Rightarrow (\sum _{j=1}^n{u}_j)(\sum _{j=1}^n{v}_j)\le n(\sum _{j=1}^n{u}_j{v}_j)$ ，这就是著名的切比雪夫单调不等式(Chebyshev's monotonic inquality)。

33. [M30] One evening Dr. Matrix discovered some formulas that might even be classed as more remarkable than those of exercise 20:

1(a−b)(a−c)+1(b−a)(b−c)+1(c−a)(c−b)=0," role="presentation" style="position: relative;">1(a−b)(a−c)+1(b−a)(b−c)+1(c−a)(c−b)=0,$\frac{1}{(a-b)(a-c)}+\frac{1}{(b-a)(b-c)}+\frac{1}{(c-a)(c-b)}=0,$

a2(a−b)(a−c)+b2(b−a)(b−c)+c2(c−a)(c−b)=1," role="presentation" style="position: relative;">a2(a−b)(a−c)+b2(b−a)(b−c)+c2(c−a)(c−b)=1,$\frac{a^2}{(a-b)(a-c)}+\frac{b^2}{(b-a)(b-c)}+\frac{c^2}{(c-a)(c-b)}=1,$

Prove that these formulas are a special case of a general law; let x1,x2,...,xn" role="presentation" style="position: relative;">x1,x2,...,xn$x_1,x_2,...,x_n$ be distinct numbers, and show that

∑j=1n(xjr/∏1≤k≤n,k≠j(xj−xk))={0,if0≤r<n−1;1,ifr=n−1;∑j=1nxj,ifr=n." role="presentation" style="position: relative;">∑nj=1(xrj/∏1≤k≤n,k≠j(xj−xk))=⎧⎩⎨⎪⎪0,1,∑nj=1xj,ififif0≤r<n−1;r=n−1;r=n.$\sum _{j=1}^n(x_j^r/\prod _{1\le k\le n,k\ne j}(x_j-x_k))=\{\begin{array}{ccc}0,& if& 0\le r<n-1;\\ 1,& if& r=n-1;\\ \sum _{j=1}^n{x}_j,& if& r=n.\end{array}$

证：

对于 0≤r≤n−1" role="presentation" style="position: relative;">0≤r≤n−1$0\le r\le n-1$ 的情况的证明如下：

∑j=1n(xjr/∏1≤k≤n,k≠j(xj−xk))" role="presentation" style="position: relative;">∑nj=1(xrj/∏1≤k≤n,k≠j(xj−xk))$\sum _{j=1}^n(x_j^r/\prod _{1\le k\le n,k\ne j}(x_j-x_k))$ 可改写为两个和式的差：1xn−xn−1(∑j=1nxjr(xj−xn−1)∏1≤k≤n,k≠j(xj−xk)−∑j=1nxjr(xj−xn)∏1≤k≤n,k≠j(xj−xk))" role="presentation" style="position: relative;">1xn−xn−1(∑nj=1xrj(xj−xn−1)∏1≤k≤n,k≠j(xj−xk)−∑nj=1xrj(xj−xn)∏1≤k≤n,k≠j(xj−xk))$\frac{1}{x_n-x_{n-1}}(\sum _{j=1}^n\frac{x_j^r(x_j-x_{n-1})}{\prod _{1\le k\le n,k\ne j}(x_j-x_k)}-\sum _{j=1}^n\frac{x_j^r(x_j-x_n)}{\prod _{1\le k\le n,k\ne j}(x_j-x_k)})$ ，每个和式都和要证明的和式是同样的形式，但都只包含 n-1 项（第一个和式少了 j=n-1 时那项，第二个少了 j=n 时那项），这就启发我们对 n 使用数学归纳法来证明。

Base case: n = 3 时，根据前面的公式显然有

1(a−b)(a−c)+1(b−a)(b−c)+1(c−a)(c−b)=0," role="presentation" style="position: relative;">1(a−b)(a−c)+1(b−a)(b−c)+1(c−a)(c−b)=0,$\frac{1}{(a-b)(a-c)}+\frac{1}{(b-a)(b-c)}+\frac{1}{(c-a)(c-b)}=0,$

 

a2(a−b)(a−c)+b2(b−a)(b−c)+c2(c−a)(c−b)=1," role="presentation" style="position: relative;">a2(a−b)(a−c)+b2(b−a)(b−c)+c2(c−a)(c−b)=1,$\frac{a^2}{(a-b)(a-c)}+\frac{b^2}{(b-a)(b-c)}+\frac{c^2}{(c-a)(c-b)}=1,$ 要证明的公式成立。

当 n≥4" role="presentation" style="position: relative;">n≥4$n\ge 4$ 时，由前面 ∑j=1n(xjr/∏1≤k≤n,k≠j(xj−xk))" role="presentation" style="position: relative;">∑nj=1(xrj/∏1≤k≤n,k≠j(xj−xk))$\sum _{j=1}^n(x_j^r/\prod _{1\le k\le n,k\ne j}(x_j-x_k))$ 可以改写为两个同样形式的 n-1 项和式的差，显然可以从 n-1 项时成立推导出 n 项时公式成立。

最后是比较麻烦的 r = n 时的情况。

由于对任意一个特定的 j（1≤j≤n" role="presentation" style="position: relative;">1≤j≤n$1\le j\le n$） ，∏k=1n(xj−xk)" role="presentation" style="position: relative;">∏nk=1(xj−xk)$\prod _{k=1}^n(x_j-x_k)$ 都等于零，因此有等式 0=∑j=1n∏k=1n(xj−xk)∏1≤k≤n,k≠j(xj−xk)=∑j=1nxjn−(x1+...+xn)xjn−1+P(xj)∏1≤k≤n,k≠j(xj−xk)" role="presentation" style="position: relative;">0=∑nj=1∏nk=1(xj−xk)∏1≤k≤n,k≠j(xj−xk)=∑nj=1xnj−(x1+...+xn)xn−1j+P(xj)∏1≤k≤n,k≠j(xj−xk)$0=\sum _{j=1}^n\frac{\prod _{k=1}^n(x_j-x_k)}{\prod _{1\le k\le n,k\ne j}(x_j-x_k)}=\sum _{j=1}^n\frac{x_j^n-(x_1+...+x_n)x_j^{n-1}+P(x_j)}{\prod _{1\le k\le n,k\ne j}(x_j-x_k)}$ ，其中的 P(xj)" role="presentation" style="position: relative;">P(xj)$P(x_j)$ 是关于 xj" role="presentation" style="position: relative;">xj$x_j$ 的最高次幂为 xjn−2" role="presentation" style="position: relative;">xn−2j$x_j^{n-2}$ 的多项式。这样可以得出 ∑j=1nxjn∏1≤k⩽n,k≠j(xj−xk)=∑j=1n(x1+...+xn)xjn−1∏1≤k⩽n,k≠j(xj−xk)+∑j=1nP(xj)∏1≤k⩽n,k≠j(xj−xk)" role="presentation" style="position: relative;">∑nj=1xnj∏1≤k⩽n,k≠j(xj−xk)=∑nj=1(x1+...+xn)xn−1j∏1≤k⩽n,k≠j(xj−xk)+∑nj=1P(xj)∏1≤k⩽n,k≠j(xj−xk)$\sum _{j=1}^n\frac{x_j^n}{\prod _{1\le k⩽n,k\ne j}(x_j-x_k)}=\sum _{j=1}^n\frac{(x_1+...+x_n)x_j^{n-1}}{\prod _{1\le k⩽n,k\ne j}(x_j-x_k)}+\sum _{j=1}^n\frac{P(x_j)}{\prod _{1\le k⩽n,k\ne j}(x_j-x_k)}$ 而根据前面已证明的结论，∑j=1nP(xj)∏1≤k⩽n,k≠j(xj−xk)=0" role="presentation" style="position: relative;">∑nj=1P(xj)∏1≤k⩽n,k≠j(xj−xk)=0$\sum _{j=1}^n\frac{P(x_j)}{\prod _{1\le k⩽n,k\ne j}(x_j-x_k)}=0$ , （因为对所有 0≤r⩽n−2" role="presentation" style="position: relative;">0≤r⩽n−2$0\le r⩽n-2$ ，∑j=1nxjr∏1≤k≤n,k≠j(xj−xk)" role="presentation" style="position: relative;">∑nj=1xrj∏1≤k≤n,k≠j(xj−xk)$\sum _{j=1}^n\frac{x_j^r}{\prod _{1\le k\le n,k\ne j}(x_j-x_k)}$ 的值都为零。）而 ∑j=1n(x1+...+xn)xjn−1∏1≤k≤n,k≠j(xj−xk)=x1+...+xn" role="presentation" style="position: relative;">∑nj=1(x1+...+xn)xn−1j∏1≤k≤n,k≠j(xj−xk)=x1+...+xn$\sum _{j=1}^n\frac{(x_1+...+x_n)x_j^{n-1}}{\prod _{1\le k\le n,k\ne j}(x_j-x_k)}=x_1+...+x_n$ ，这样就证明了 r=n 时的情况：∑j=1nxjn∏1≤k≤n,k≠j(xj−xk)=x1+...+xn" role="presentation" style="position: relative;">∑nj=1xnj∏1≤k≤n,k≠j(xj−xk)=x1+...+xn$\sum _{j=1}^n\frac{x_j^n}{\prod _{1\le k\le n,k\ne j}(x_j-x_k)}=x_1+...+x_n$ 。

34. [M25] 证明如下等式成立：

 假定其中  且 x" role="presentation" style="position: relative;">x$x$ 为任意实数。例如，当 n=4" role="presentation" style="position: relative;">n=4$n=4$ 且 m=2" role="presentation" style="position: relative;">m=2$m=2$ 时，x(x−2)(x−3)(−1)(−2)(−3)+(x+1)(x−1)(x−2)(1)(−1)(−2)+(x+2)x(x−1)(2)(1)(−1)+(x+3)(x+1)x(3)(2)(1)=1" role="presentation" style="position: relative;">x(x−2)(x−3)(−1)(−2)(−3)+(x+1)(x−1)(x−2)(1)(−1)(−2)+(x+2)x(x−1)(2)(1)(−1)+(x+3)(x+1)x(3)(2)(1)=1$\frac{x(x-2)(x-3)}{(-1)(-2)(-3)}+\frac{(x+1)(x-1)(x-2)}{(1)(-1)(-2)}+\frac{(x+2)x(x-1)}{(2)(1)(-1)}+\frac{(x+3)(x+1)x}{(3)(2)(1)}=1$

证：

和33题的公式对照一下就可以找出类似的模式：令xk=k,xr=r" role="presentation" style="position: relative;">xk=k,xr=r$x_k=k,x_r=r$ ，分子 ∏1≤r≤n,r≠m(x+k−r)=kn−1+P(k)" role="presentation" style="position: relative;">∏1≤r≤n,r≠m(x+k−r)=kn−1+P(k)$\prod _{1\le r\le n,r\ne m}(x+k-r)=k^{n-1}+P(k)$ ，其中的 P(k)" role="presentation" style="position: relative;">P(k)$P(k)$ 是最高次幂为  的关于 k" role="presentation" style="position: relative;">k$k$ 的多项式，套用 33 题的公式证明过程，∑k=1nP(k)∏1≤r≤n,r≠k(k−r)=0" role="presentation" style="position: relative;">∑nk=1P(k)∏1≤r≤n,r≠k(k−r)=0$\sum _{k=1}^n\frac{P(k)}{\prod _{1\le r\le n,r\ne k}(k-r)}=0$ ，这样就可以直接得出 ∑k=1n∏1≤r≤n,r≠m(x+k−r)∏1≤r≤n,r≠m(k−r)=∑k=1nkn−1∏1≤r≤n,r≠m(k−r)=1" role="presentation" style="position: relative;">∑nk=1∏1≤r≤n,r≠m(x+k−r)∏1≤r≤n,r≠m(k−r)=∑nk=1kn−1∏1≤r≤n,r≠m(k−r)=1$\sum _{k=1}^n\frac{\prod _{1\le r\le n,r\ne m}(x+k-r)}{\prod _{1\le r\le n,r\ne m}(k-r)}=\sum _{k=1}^n\frac{k^{n-1}}{\prod _{1\le r\le n,r\ne m}(k-r)}=1$​​​​​​​