二项式定理 Hurwitz的推广

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ProofWiki
$$(x + y)^n = \sum x (x + \epsilon_1 z_1 + \cdots + \epsilon_n z_n)^{\epsilon_1 + \cdots + \epsilon_n - 1}(y - \epsilon_1 z_1 - \cdots - \epsilon_n z_n)^{n - \epsilon_1 - \cdots - \epsilon_n}$$
where the summation ranges over all $2^n$ choices of $\epsilon_1, \ldots, \epsilon_n = 0$ or $1$ independently.

该网站没有给出证明..如何证明？

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For $n=1$, the formula is
$(x + y)^1 = x(x + 0z_1)^{0-1}(y - 0z_1)^{1-0}+x(x + 1z_1)^{1-1}(y - 1z_1)^{1-1}$
For $n=2$, the formula is
\begin{align*}
(x + y)^2 &= x(x + 0z_1 + 0z_2)^{0+0-1}(y - 0z_1 - 0z_2)^{2-0-0}\\
&+ x(x + 1z_1 + 0z_2)^{1+0-1}(y - 1z_1 - 0z_2)^{2-1-0}\\
&+ x(x + 0z_1 + 1z_2)^{0+1-1}(y - 0z_1 - 1z_2)^{2-0-1}\\
&+ x(x + 1z_1 + 1z_2)^{1+1-1}(y - 1z_1 - 1z_2)^{2-1-1}
\end{align*}

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证明该公式需要证明 Knuth TAOCP 的练习 2.3.4.4-30
\[ \sum x \left(x + \epsilon_1 z_1 + \cdots + \epsilon_n z_n\right)^{\epsilon_1 + \cdots + \epsilon_n - 1} \left(y + (1 - \epsilon_1) z_1 - \cdots + \left(1 - \epsilon_n\right) z_n\right)^{n - \epsilon_1 - \cdots - \epsilon_n}= \left(x + y\right)\left(x + y + z_1 + \cdots + z_n\right)^{n - 1}\]