It is found that the even powers of the Vandermonde alternating functionare given by$beginaligned V_2nx_1x_2ldotsx_n&=prod_1leq i继续

\leq j \leq n}(x_j-x_i)^2 \ &=\sum_{1\leq i_1<i_2<\cdots<i_n\leq 2n}\prod_{1\leq j<k\leq n}(x_{i_k}-x_{i_j})^2 \end{aligned}$

Proof:

We will use induction on $n$. For $n=1$, we have $V_2(x_1,x_2)=(x_2-x_1)^2$, which is true.

Assume that the statement is true for $n-1$. We will prove it for $n$.

Using the identity $x^2-y^2=(x+y)(x-y)$, we have:

$\begin{aligned} V_{2n}(x_1,x_2,\ldots,x_n) &=\prod_{1\leq i<j\leq n}(x_j-x_i)^2 \ &=\prod_{1\leq i<j\leq n}\left[(x_j+x_i)(x_j-x_i)\right]^2 \ &=\left[\prod_{1\leq i<j\leq n}(x_j+x_i)^2\right]\left[\prod_{1\leq i<j\leq n}(x_j-x_i)^2\right] \ &=\left[\prod_{1\leq i<j\leq n}(x_j+x_i)^2\right]V_{2(n-1)}(x_1,x_2,\ldots,x_{n-1}) \end{aligned}$

Now, we expand the product $\prod_{1\leq i<j\leq n}(x_j+x_i)^2$ using the induction hypothesis:

$\begin{aligned} \prod_{1\leq i<j\leq n}(x_j+x_i)^2 &=\prod_{1\leq i<j\leq n-1}(x_j+x_i)^2(x_n+x_i)^2(x_n+x_j)^2 \ &=\sum_{1\leq i_1<i_2<\cdots<i_{n-1}\leq 2(n-1)}\prod_{1\leq j<k\leq n-1}(x_{i_k}+x_{i_j})^2(x_{i_k}+x_n)^2(x_{i_j}+x_n)^2(x_{i_k}+x_{i_j})^2(x_{i_k}-x_{i_j})^2 \end{aligned}$

Substituting this in the expression for $V_{2n}(x_1,x_2,\ldots,x_n)$, we get:

$\begin{aligned} V_{2n}(x_1,x_2,\ldots,x_n) &=\sum_{1\leq i_1<i_2<\cdots<i_{n-1}\leq 2(n-1)}\prod_{1\leq j<k\leq n-1}(x_{i_k}+x_{i_j})^2(x_{i_k}+x_n)^2(x_{i_j}+x_n)^2(x_{i_k}+x_{i_j})^2(x_{i_k}-x_{i_j})^2V_{2(n-1)}(x_1,x_2,\ldots,x_{n-1}) \ &=\sum_{1\leq i_1<i_2<\cdots<i_{n}\leq 2n}\prod_{1\leq j<k\leq n}(x_{i_k}-x_{i_j})^2 \end{aligned}$

where the last equality follows from the fact that the summands in the second line are precisely the terms in the expansion of the product in the first line, with $i_n=2n$ added to the indices.

Therefore, the statement is true for all $n\geq 1$ by induction.