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\(\binom{n}{k}k^{\_{m}} = \binom{n-m}{k-m}n^{\_{m}}\)

证明过程：
\( \begin{equation} \begin{split} \binom{n}{k}k^{\_{m}} &= \frac{n!}{k!(n-k)!}k(k-1)(k-2)……(k-m+1)\\ &=\frac{n!}{(k-m)!(n-k)!}\\ &=\frac{(n-m)!}{(k-m)!((n-m)-(k-m))!}n^{\_{m}}\\ &=\binom{n-m}{k-m}n^{\_{m}} \end{split} \end{equation} \)

\( \begin{aligned} \frac{((\sum_{i=1}^{n}x_i \times \overline{x})+\overline{x})^2}{\overline{x}^2} &= \frac{(\overline{x} \times \sum_{i=1}^{n}x_i)^2+2(\overline{x} \times \sum_{i=1}^{n}x_i)\overline{x}+\overline{x}^2}{\overline{x}^2}\\ &=\frac{\overline{x}^2(\sum_{i=1}^{n}x_i)^2}{\overline{x}^2}+\frac{2\overline{x}(\sum_{i=1}^{n}x_i)}{\overline{x}}+1\\ &=(\sum_{i=1}^{n}x_i)^2+2(\sum_{i=1}^{n}x_i)+1 \end{aligned} \)