Let $A_{k}=\{0,... ,n\}\setminus\{k\}.$ How to prove $\sum_{k=0}^{n}\left[(-1)^{k+1}\prod_{\substack{i,j\in A_{k}\\i

Question

Let $A_{k}=\{0,1,\ldots,n\}\setminus\{k\}$ for each $k=0,1,\ldots ,n$.

I think that the following equality is true for all $n\in\mathbb{N}, n\geq 2$ : \begin{align} \sum_{k=0}^{n}\left[(-1)^{k+1}\prod_{\substack{i,j\in A_{k}\\i<j}}(a_{i}-a_{j})\right]=0. \end{align} But I cannot prove it so that I want the proof.

$\displaystyle\prod_{\substack{i,j\in A_{k}\\i<j}}(a_{i}-a_{j})\quad$is equal to, for example, \begin{align} &(a_{1}-a_{2})\quad&\mathrm{if}\quad n=2,k=0,\\ &(a_{1}-a_{2})(a_{1}-a_{3})(a_{2}-a_{3})\quad&\mathrm{if}\quad n=3,k=0,\\ &(a_{1}-a_{2})(a_{1}-a_{3})(a_{1}-a_{4})(a_{2}-a_{3})(a_{2}-a_{4})(a_{3}-a_{4})\quad&\mathrm{if}\quad n=4,k=0. \end{align} Thus, we get this product by multiplying ${}_{n}\mathrm{C}_{2}$ differences.

If $n=2$, \begin{align} \sum_{k=0}^{n}\left[(-1)^{k+1}\prod_{\substack{i,j\in A_{k}\\i<j}}(a_{i}-a_{j})\right]=-(a_{1}-a_{2})+(a_{0}-a_{2})-(a_{0}-a_{1})=0. \end{align} To show that the equality is true for all $n\in\mathbb{N}, n\geq 2,$

should I use mathematical induction, or is there another good method for proving it?

Answer 1

For pairwise distinct $a_0, \ldots, a_n$ is $$ \prod_{\substack{i,j\in A_{k}\\i<j}}(a_{i}-a_{j}) = \frac{\prod\limits_{0 \le i < j \le n }(a_i - a_j)}{\prod\limits_{0 \le i < k}^n(a_i-a_k) \prod\limits_{k < j \le n}^n(a_k-a_j)} = \frac{\prod\limits_{0 \le i < j \le n }(a_i - a_j)}{(-1)^{n-k}\prod\limits_{\substack{i=0 \\ i \ne k}}^n (a_i - a_k)} $$ so that your expression is equal to $$ (-1)^{n+1} \cdot\prod\limits_{0 \le i < j \le n }(a_i - a_j) \cdot \sum_{k=0}^n \frac{1}{\prod\limits_{\substack{i=0 \\ i \ne k}}^n (a_i - a_k)} $$ and that is zero, see for example Showing that $\sum_{i = 1}^m \frac{1}{\prod_{j = 1, j \neq i}^m (a_j - a_i)}$ is zero.

Since your expression is a continuous function of $a_0, \ldots, a_n$ the result still holds if the numbers are not pairwise distinct.