Vandermonde行列式的一个推广

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“Vandermonde行列式的一个推广及其在初等数学中的应用” 肖振纲
文〔1〕、〔2〕先后利用复变函数中的留数定理讨论了初等数学中的I.J.Matrix定理的两个推广, 文〔3〕则用初等方法讨论了所述两个推广的关系。
本文首先给出高等代数中著名的Vandermonde行列式^〔4〕的一个推广(不同于文〔5〕),然后简单地导出I.J.Matrix定理的两个推广, 并由此进一步给出几类新颖而有趣的组合恒等式.
定理1 设 $n$ 是一个正整数, $r$ 是一个非负整数,对任意 $n+1$ 个数 $x_{0}, x_{1}, \cdots, x_n$, 令
$$\tag1\label1
D(n, r)=\left|\begin{array}{cccccc}
1 & x_{0} & x_{0}^{2} & \cdots & x_{0}^{n-1} & x_{0}^{r} \\
1 & x_{1} & x_{1}^{2} & \cdots & x_{1}^{n-1} & x_{1}^{r} \\
\cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\
1 & x_{n} & x_{n}^{2} & \cdots & x_{n}^{n-1} & x_{n}^{r}
\end{array}\right|
$$
则有$$\tag2\label2 D(n,r)=\begin{cases}0&,(0\le r\le n-1)\\\prod_{0\le j<i\le n}(x_i-x_j)\sum_{j_{0}+j_{1}+\cdots+j_n=r-n} x_{0}^{j_{0}}x_{1}^{j_{1}} \cdots  x_n^{j_n}&,(r\ge n)\end{cases}$$
其中, $j_{0}, j_{1}, \cdots, j_{n}$皆为非负整数 (在后面的讨论中, 这一说明从略)。
证明 当 $0 \leqslant r \leqslant n-1$ 时, $D(n, r)$ 中有两列元素相同, 所以, $D(n, r)=0$,下面用数学归纳法证明 $r \geqslant n$ 的情形。
当 $r \geqslant 1$ 时, 因
$$D(1, r)=x_{1}^{r}-x_{0}^{r}=\left(x_{1}-x_{0}\right)\left(x_{0}^{r-1}+x_0^{r-2} x_{1}+\cdots+x_{1}^{r-1}\right)=\left(x_{1}-x_{0}\right) \sum_{j_{0}+j_{1}=r-1} x_0^{j_{0}} x_{1}^{j_{1}}$$所以, 当 $n=1$ 时, 结论成立, 设结论对 $n-1(n \geqslant 2, r \geqslant n-1)$ 的情形成立, 现考虑 $n(r \geqslant n)$ 的情形。
将行列式 $D(n, r)$ 中的第 $n+1$ 列减去第 $n$ 列的$x_{n}^{r-n+1}$ 倍, 而对其余的列, 自右至左依次从每一列减去它前一列的 $x_n$ 倍,则由行列式的性质,有$$\eqalign{D(n,r)&=\begin{vmatrix}
1 & x_{0}-x_{n} & x_{0}^{2}-x_{0} x_n&⋯& x_{0}^{n-1}-x_{0}^{n-2} x_n & x_{0}^{r}-x_{0}^{n-1} x_n^{r-n+1}\\
1 & x_{1}-x_{n} & x_{1}^{2}-x_{1} x_{n}&⋯& x_{1}^{n-1}-x_{1}^{n-2} x_{n} & x_{1}^{r}-x_{1}^{n-1} x_n^{r-n+1}\\
⋯ & ⋯ & ⋯ & ⋯ &⋯& ⋯\\
1 & x_{n-1}-x_n& x_{n-1}^{2}-x_{n-1} x_n &⋯& x_{n-1}^{n-1}-x_{n-1}^{n-2} x_{n} & x_{n-1}^r-x_{n-1}^{n-1} x_{n}^{r-n+1}\\
1 & 0 & 0 &⋯& 0 & 0\end{vmatrix}\\
&=(-1)^{n+2}\begin{vmatrix}
x_{0}-x_n & x_{0}^{2}-x_{0} x_n& \cdots & x_{0}^{n-1}-x_{0}^{n-2} x_n& x_{0}^{r}-x_{0}^{n-1} x_n^{r-n+1}\\
x_{1}-x_n & x_{1}^{2}-x_{1} x_n & \cdots& x_{1}^{n-1}-x_{1}^{n-2} x_n& x_{1}^{r}-x_{1}^{n-1} x_n^{r-n+1}\\
⋯ & ⋯ & ⋯ & ⋯ & ⋯\\
x_{n-1}-x_n&x_{n-1}^{2}-x_{n-1} x_n & \cdots & x_{n-1}^{n-1}-x_{n-1}^{n-2} x_n & x_{n-1}^{r}-x_{n-1}^{n-1} x_n^{r-n+1}\end{vmatrix}\\
&=\prod_{i=0}^{n-1}\left(x_n-x_{i}\right)\begin{vmatrix}
1 & x_{0} & \cdots & x_{0}^{n-2} & x_{0}^{n-1} \sum_{k+j_n=r-n} x_{0}^{k} x_n^{j_{n}}\\
1 & x_{1} & \cdots & x_{1}^{n-2} & x_{1}^{n-1} \sum_{k+j_{n}=r-n} x_{1}^{k} x_{n}^{j_n}\\
⋯ & ⋯ & ⋯ & ⋯ & ⋯\\
1 & x_{n-1} & \cdots & x^{n-2}_{n-1} & x_{n-1}^{n-1} \sum_{k+j_n=r-n} x_{n-1}^{k} x_{n}^{j_{n}}
\end{vmatrix}\\
&=\prod_{i=0}^{n-1}\left(x_n-x_{i}\right) \sum_{k+j_n=r-n} x_{n}^{j_n}\left|\begin{array}{c}1 & x_{0} & \cdots & x_{0}^{n-2} & x_{0}^{n-1+k} \\ 1 & x_{1} & \cdots & x_{1}^{n-2} & x_{1}^{n-1+k} \\ \cdots & \cdots & \cdots & \cdots & \cdots \\ 1 & x_{n-1} & \cdots & x_{n-1}^{n-2} & x_{n-1}^{n-1+k}\end{array}\right|\\
&=\prod_{i=0}^{n-1}\left(x_n-x_{i}\right)\sum_{k+j_n=r-n}x_n^{j_n}D(n-1,n-1+k)}$$因$n-1+k\geq n-1$,由归纳假设,有
$$D(n-1,n-1+k)=\prod_{0\le j<i\le n-1}(x_i-x_j)\sum_{j_{0}+j_{1}+\cdots+j_{n-1}=k} x_{0}^{j_{0}} x_{1}^{j_{1}} \cdots \cdots x_{n-1}^{j_{n-1}}$$于是, 当$r\geqslant n$时, 有$$\eqalign{D(n,r)&=\prod_{i=0}^{n-1}\left(x_n-x_{i}\right)\prod_{0\le j<i\le n-1}\left(x_{i}-x_{j}\right)\sum_{k+j_n=r-n}x_n^{j_n}\sum_{j_0+j_1+\dots+j_{n-1}=k}x_0^{j_0}x_1^{j_1}\dots x_{n-1}^{j_{n-1}}
\\&=\prod_{0\le j<i\le n}\left(x_{i}-x_{j}\right) \sum_{j_{0}+j_{1}+\cdots+j_n=r-n} x_{0}^{j_{0}} x_{1}^{j_1} \cdots x_n^{j_n}}$$因此，结论对$n$的情形也成立，故结论对一切正整数皆成立，证毕。
特别地，当$r=n$时，$D(n,n)$即为$n+1$阶Vandermonde行列式，由于$$\sum_{j_{0}+j_{1}+\cdots+j_n=0} x_0^{j_0} x _{1}^{j_{1}} \cdots x_n^{j_{n}}=1$$因此,$$D(n, n)=\prod_{0 \leqslant j<i \leqslant n}\left(x_{i}-x_{j}\right)$$这正是著名的Vandemonde行列式的结论, 故定理1是它的一个推广.
我们还可以将$r$拓广到负整数.
定理2 设$n,r$皆为正整数, $x_0,x_1,\dots,x_{n}$是$n+1$个不全为零的数, 记$$\tag3\label3d=\left|\begin{array}{c}1 & x_{0} & x_{0}^2& \cdots & x_{0}^{n-1} & x_{0}^{-r} \\ 1 & x_{1} & x_{1}^2 & \cdots & x_{1}^{n-1} & x_{1}^{-r} \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ 1 & x_n & x_n^2 & \cdots & x_n^{n-1} & x_n^{-r}\end{array}\right|$$则$$\tag4\label4d=\frac{(-1)^n \prod_{0 \leqslant j<i \leqslant n}\left(x_{i}-x_{j}\right)}{\prod_{0\leqslant i\leqslant n}x_i} \sum_{j_{0}+j_{1}+\cdots+j_n=r-1} x_{0}^{-j_{0}} x_{1}^{-j_{1}} \cdots x_n^{-j_n}$$证明$$\eqalign{d=&\left(\prod_{0 \leqslant i \leqslant n} x_{i}\right)^{n-1}\left|\begin{array}{ccccc}x_{0}^{-(n-1)} & x_{0}^{-(n-2)} & \cdots & 1 & x_{0}^{-(r+n-1)} \\ x_{1}^{-(n-1)} & x_{1}^{-(n-2)} & \cdots & 1 & x_{1}^{-(r+n-1)} \\ \cdots & \cdots & \cdots & \cdots & \cdots \\ x_{n}^{-(n-1)} & x_{n}^{-(n-2)} & \cdots & 1 & x_{n}^{-(r+n-1)}\end{array}\right|\quad各行提取x_i^{n-1}\\
=&(-1)^{\frac{n(n-1)}{2}}\left(\prod_{0 \leq i \leq n} x_{i}\right)^{n-1}\left|\begin{array}{ccccc}1 & x_{0}^{-1} & \cdots & x_{0}^{-(n-1)} & x_{0}^{-(r+n-1)} \\ 1 & x_{1}^{-1} & \cdots & x_{1}^{-(n-1)} & x_{1}^{-(r+n-1)} \\ \cdots & \cdots & \cdots & \cdots & \cdots \\ 1 & x_n^{-1} & \cdots & x_{n}^{-(n-1)} & x_n^{-(r+n-1)}\end{array}\right|\quad将前n列倒过来\\
=&(-1)^{\frac{n(n-1)}{2}}\left(\prod_{0 \leqslant i \leqslant n} x_{i}\right)^{n-1} \prod_{0 \leqslant j<i \leqslant n}\left(x_{i}^{-1}-x_{j}^{-1} \right)\sum_{j_{0}+j_{1}+\cdots+j_{n}=r-1} x_0^{-j_{0}} x_{1}^{-j_{1}} \cdots x_{n}^{-j_{n}}
\quad由定理1}$$但$$\prod_{0 \leqslant j<i \leqslant n}\left(x_{i}^{-1}-x_{j}^{-1}\right)=\left(\prod_{0 \leqslant i \leqslant n} x_{i}\right)^{-n} \prod_{0 \leqslant j<i \leqslant n}\left(x_{j}-x_{i}\right)=(-1)^{\frac{n(n+1)}{2}}\left(\prod_{0 \leqslant i \leqslant n} x_{i}\right)^{-n} \prod_{0 \leqslant j<i \leqslant n}\left(x_{i}-x_{j}\right)$$故$$\eqalign{d&=(-1)^{n^{2}}\left(\prod_{0 \leqslant i \leqslant n} x_{i}\right)^{-1} \prod_{0 \leqslant j<i \leqslant n}\left(x_{i}-x_{j}\right)\sum_{j_{0}+j_{1}+\cdots+j_{n}=r-1} x_0^{-j_0} x_{1}^{-j_1} \cdots x_{n}^{-j_n}\\&=\frac{(-1)^n\prod_{0 \leqslant j<i \leqslant n}\left(x_{i}-x_{j}\right)}{\prod_{0 \leqslant i \leqslant n} x_{i}}\sum_{j_{0}+j_{1}+\cdots+j_n=r-1} x_0^{-j_0} x_{1}^{-j_1} \cdots x_n^{-j_n}\quad\text{证毕}}$$
Vandermonde行列式的一个推广及其在初等数学中的应用_肖振纲.pdf (352.04 KB, Downloads: 46)

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现在给出推广后的Vandermonde行列式在初等数学中的一些应用。
设$x_0,x_1,\dots,x_n$互不相等. 记$$A_{k}=\left|\begin{array}{ccccc}1 & x_{0} & x_{0}^{2} & \cdots & x_{0}^{n-1} \\ \cdots & \cdots & \cdots & \cdots & \cdots \\ 1 & x_{k-1} & x_{k-1}^{2} & \cdots & x_{k-1}^{n-1} \\ 1 & x_{k+1} & x_{k+1}^2 & \cdots & x_{k+1}^{n-1} \\ \cdots & \cdots & \cdots & \cdots & \cdots \\ 1 & x_n & x_n^{2} & \cdots & x_n^{n-1}\end{array}\right|$$其中$k=0,1,2,\dots,n$. 则$A_k$是一个$n$阶Vandermonde行列式,因此有$$A_k=\prod_{\substack{0\le j<i\le n\\i,j\ne k}}(x_i-x_j)=(-1)^{n-k}\frac{\prod_{0\le j<i\le n}(x_i-x_j)}{\prod_{\substack{0\le j\le n\\j\ne k}}(x_k-x_j)}$$于是，将行列式\eqref{1}按第$n+1$列展开即得$$D(n, r)=\sum_{k=0}^n(-1)^{n+k} x_{k}^{r} A_{k}=\prod_{0 \leqslant j<i \leqslant n}\left(x_{i}-x_{j}\right)\sum_{k=0}^{n} \frac{x_{k}^r}{\prod_{\substack{0\leqslant j \leqslant n \\ j \neq k}}\left(x_{k}-x_{j}\right)}$$从而由\eqref{2}式立即得到
定理3 设$x_0,x_1,…,x_n$,是$n+1$个互不相等的数，$r$是一个非负整数，则有$$\tag5\label5\sum_{k=0}^{n} \frac{x_{k}^r}{\prod_{\substack{0\leqslant j \leqslant n \\ j \neq k}}\left(x_{k}-x_{j}\right)}=\left\{\begin{array}{l}0&(0\leqslant r\leqslant n-1)\\ \sum_{j_{0}+j_{1}+\cdots+j_{n}=r-n} x_0^{j_0} x_{1}^{j_{1}} \cdots x_{n}^{j_{n}}&(r \geqslant n)\end{array}\right.$$\eqref{5}式即文〔1〕所述I・J・Matrix定理的推广。
同样，将行列式\eqref{3}按第$n+1$列展开，再由\eqref{4}式可得。
定理4 设$x_0,x_1,…,x_n$是$n+1$个互不相等且全不为零的数，$r$是一个正整数，则有$$\tag6\label6\sum_{k=0}^{n} \frac{x_{k}^{-r}}{\prod_{\substack{0\leqslant j \leqslant n \\ j \neq k}}\left(x_{k}-x_{j}\right)}=\frac{(-1)^{n}}{\prod_{0 \leqslant i \leqslant n} x_{i}} \sum_{j_{0}+j_{1}+\cdots+j_n=r-1} x_{0}^{-j_{0}} x_{1}^{-j_1} \cdots x_{n}^{-j_{n}}$$如果$x_0,x_1,…,x_n$是一个公差为$d(≠0)$的等差数列，则易知$$\frac{1}{\prod_{\substack{0\leqslant j\leqslant n \\ j \neq k}}\left(x_{k}-x_{j}\right)}=(-1)^{n-k} \frac{C_n^{k}}{n ! d^{n}}$$其中，$k=0,1,2,⋯,n$.
于是由定理3即可得到一类与等差数列有关的深刻而有趣的组合恒等式：
定理5 设$a_0,a_1,…,a_n$是一个公差为$d$的等差数列，$r$是一个非负整数，则有$$\tag7\label7\sum_{k=0}^{n}(-1)^{n-k} a_k^rC_{n}^{k}=\left\{\begin{array}{ll}0 & (0 \leqslant r \leqslant n-1) \\ n ! d^n \sum_{j_{0}+j_{1}+\cdots+j_{n}=r-n} a_{0}^{j_0} a_{1}^{j_{1}} \cdots a_n^{j_{n}}&(r \geqslant n)\end{array}\right.$$注意到$j_0,j_1,⋯,j_n$都取非负整数时，有$\sum_{j_{0}+j_{1}+\cdots+j_n=0} a_{0}^{j_{0}} a_{1}^{j_{1}} \cdots \cdot a_n^{j_n}=1$
$$\sum_{j_{0}+j_{1}+\cdots+j_n=1} a_{0}^{j_{0}} a_{1}^{j_{1}} \cdots a_n^{j_{n}}=\sum_{i=0}^n a_{i}$$
$$\sum_{j_{0}+j_{1}+\cdots+j_{n}=2} a_{0}^{j_{0}} a_{1}^{j_{1}} \cdots a_n^{j_{n}}=\sum_{i=0}^na_{i}^{2}+\sum_{0 \leqslant i \leqslant j \leqslant n} a_{i} a_{j}=\frac{1}{2} \sum_{i=0}^{n} a_{i}^{2}+\frac{1}{2}\left(\sum_{i=0}^na_{i}\right)^{2}$$
$$\eqalign{\sum_{j_{0}+j_{1}+\cdots+j_{n}=3} a_{0}^{j_{0}} a_{1}^{j_{1}} \cdots a_n^{j_{n}}&=\sum_{i=0}^{n} a_{i}^{3}+\sum_{i \neq j} a_{i}^{2} a_{j}+\sum_{0\leqslant i<j<k \leqslant n} a_{i} a_{j} a_k\\&=\frac{1}{3} \sum_{i=0}^{n} a_{i}^{3}+\frac{1}{2}\left(\sum_{i=0}^n a_{i}^{2}\right)\left(\sum_{i=0}^{n} a_{i}\right)+\frac{1}{6}\left(\sum_{i=0}^{n} a_{i}\right)^3}$$
又当$a_0,a_1,…,a_n$是公差为$d$的等差数列时，有
$$\tag8\label8\sum_{i=0}^n a_{i}=\frac{1}{2}(n+1)\left(a_{0}+a_n\right)$$
$$\tag9\label9\sum_{i=0}^n a_{i}^{2}=\frac{1}{6} n(n+1) d^{2}+\frac{1}{3}(n+1)\left(a_{0}^{2}+a_{0} a_{n}+a_{n}^{2}\right)$$
$$\tag{10}\label{10}\sum_{i=0}^n a_{i}^{3}=\frac{1}{4}(n+1)\left(a_{0}+a_n\right)\left(a_{0}^{2}+a_n^{2}+n d^{2}\right)$$
其中，\eqref{8}式是熟知的，\eqref{9}\eqref{10}两式不难用数学归纳法或其它方法证明，于是可得
$\sum_{j_{0}+j_{1}+\cdots+j_n=1} a_{0}^{j_{0}} a_{1}^{j_{1}} \cdots a_n^{j_{n}}=\frac{1}{2}(n+1)\left(a_{0}+a_n\right)$
$\sum_{j_0+j_1+\cdots+j_n=2}a_0^{j_0}a_1^{j_1}\cdots a_n^{j_n}=\frac1{12}(n+1)\left[(3n+5)\left(a_0+a_n\right)^2+2\left(a_0^2+a_n^2+nd^2\right)\right]$
$\sum_{j_0+j_1+\cdots+j_n=3}a_0^{j_0}a_1^{j_1}\cdots a_n^{j_n}=\frac1{48}(n+1)(n+3)\left(a_0+a_n\right)\left[(n+1)\left(a_0+a_n\right)^2+2\left(a_0^2+a_n^2+nd^2\right)\right]$
故在\eqref{7}式中分别令$r=n,n+1,n+2,n+3$即得
推论 设$a_0,a_1,…,a_n$是一个公差为$d$的等差数列，则有
$$\label{11}\tag{11}\sum_{k=0}^n(-1)^{n-k}a_k^nC_n^k=n!d^n$$
$$\label{12}\tag{12}\sum_{k=0}^n(-1)^{n-k}a_k^{k+1}C_n^k=\frac{(n+1)!}2d^n\left(a_0+a_n\right)$$
$$\label{13}\tag{13}\sum_{k=0}^n(-1)^{n-k}a_k^{n+2}C_n^k=\frac{(n+1)!}{12}d^n\left[(3n+5)\left(a_0+a_n\right)^2+2\left(a_0^2+a_n^2+nd^2\right)\right]$$
$$\label{14}\tag{14}\sum_{k=0}^n(-1)^{n-k}a_k^{n+3}C_n^k=\frac{(n+1)!}{48}d^n(n+3)\left(a_0+a_n\right)\left[(n+1)\left(a_0+a_n\right)^2+2\left(a_0^2+a_n^2+nd^2\right)\right]$$
当$a_k=k(k=0,1,\cdots,n)$时,$d=1$,于是由推论,可得
$$\label{15}\tag{15}\sum_{k=1}^n(-1)^{n-k}k^nC_n^k=n!$$
$$\label{16}\tag{16}\sum_{k=1}^n(-1)^{n-k}k^{n+1}C_n^k=\frac12n(n+1)!$$
$$\label{17}\tag{17}\sum_{k=1}^n(-1)^{n-k}k^{n+2}C_n^k=\frac{1}{24} n(3 n+1)(n+2) !$$
$$\label{18}\tag{18}\sum_{k=1}^n(-1)^{n-k}k^{n+3}C_n^k=\frac{1}{48} n^2(n+1)(n+3) !$$
同样，由定理4可得另一类与等差数列有关的有趣的组合恒等式。
定理6 设$a_0,a_1,\dots,a_n$是一个公差为$d$的，各项都不为零的等差数列，$r$是一个正整数，则有$$\label{19}\tag{19}\sum_{k=0}^n(-1)^ka_k^{-r}C_n^k=\frac{n!d^n}{\prod_{0\le i\le n}a_i}\sum_{j_0+j_1+\dots+j_n=r-1}a_0^{-j_0}a_1^{-j_1}\dots a_n^{-j_n}$$特别地,当$r$分别为1,2时,由\eqref{19}式可得$$\label{20}\tag{20}\sum_{i=0}^{n}(-1)^{2} \frac{C_n^{k}}{a_{i}}=\frac{n ! d!}{\prod_{0<i<!} a_{i}}$$
$$\label{21}\tag{21}\sum_{i=0}^{n}(-1)^{2} \frac{C_n^{k}}{a_{i}^2}=\frac{n ! d^{!}}{\prod_{0<i<!} a_{i}}\sum_{k=0}^n\frac1{a_i}$$若取$a_k=k+1(0\le k\le n)$,则\eqref{20},\eqref{21}变为$$\label{22}\tag{22}\sum_{k=0}^{n}(-1)^{k}\frac{C_{n}^{k}}{k+1}=\frac{1}{n+1}$$
$$\label{23}\tag{23}\sum_{k=0}^{n}(-1)^{k}\frac{C_{n}^{k}}{\left(k+1\right)^{2}}=\frac{1}{n+1}\sum_{k=1}^{n+1}\frac{1}{k}$$若取$a_k=2k+1(0≤k≤n)$,则由\eqref{20}\eqref{21}两式得(注意$(2n+1)!=2^n·n!(2n+1)!!$)
$$\label{24}\tag{24}\sum_{k=0}^{n}(-1)^{k} \frac{C_{n}^{k}}{2 k+1}=\frac{4^{n}}{(2 n+1) C_{2 n}^{n}}$$
$$\label{25}\tag{25}\sum_{k=0}^{n}(-1)^{k} \frac{C_{n}^{k}}{(2 k+1)^{2}}=\frac{4^{n}}{(2 n+1) C_{2 n}^{n}} \sum_{k=0}^{n} \frac{1}{2 k+1}$$

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式\eqref{5}当$r\le n-1$时,也可以这样证明:
由拉格朗日插值公式, 因为$f(x)=x^r$的次数$r$不超过$n-1$, 所以$f(x)$等于$f$在$n$个点处的插值多项式 [例如2次函数等于它在3个点处的插值多项式]
$$x^r=\sum_{k=0}^{n}x_{k}^r \frac{\prod_{\substack{0\leqslant j \leqslant n \\ j \neq k}}\left(x-x_{j}\right)}{\prod_{\substack{0\leqslant j \leqslant n \\ j \neq k}}\left(x_{k}-x_{j}\right)}$$右边的每一项都是$n$次多项式,所以右边的$n,n-1,\dots,r+1$次项系数之和为0. 下面分开来求:

$\hspace1.2emn$次项系数之和	$$\tag50=\sum_{k=0}^{n}x_k^r\frac1{\prod_{\substack{0≤j ≤n \\ j ≠ k}}\left(x_{k}-x_{j}\right)}$$
$n-1$次项系数之和	$$0=\sum_{k=0}^{n}x_k^r\frac{\sum_{\substack{0≤j ≤ n \\ j≠k}} x_j }{\prod_{\substack{0≤j ≤ n \\ j≠k}}\left(x_{k}-x_{j}\right)}$$
$n-2$次项系数之和	$$0=\sum_{k=0}^{n}x_k^r\frac{\sum_{\substack{0≤i<j ≤ n \\ j≠k}} x_ix_j }{\prod_{\substack{0≤j ≤ n \\ j≠k}}\left(x_{k}-x_{j}\right)}$$
$$⋯$$	$$0=\sum_{k=0}^{n}x_k^r\frac{⋯}{\prod_{\substack{0≤j ≤ n \\ j≠k}}\left(x_{k}-x_{j}\right)}$$

又见这帖

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式\eqref{6}的$r=2$情况在这帖有一个用拉格朗日插值公式的证明