子江 | 随笔

2020-01-03 本文已影响0人鹿江kae

我们想把形如
$\begin{equation} A_i = (n+i) (n+i-1) \cdots (n+2) (n+1) \end{equation}$
的式子拆分为一系列形如
$\begin{equation} B_{i,j} = n (n-1) \cdots (n-i+j+1) \end{equation}$
的式子的相加.

那么有
$\begin{equation} \begin{split} A_{i+1} = &(n+i+1) (n+i) (n+i-1) \cdots (n+2) (n+1) \\ = &(n+i+1) A_i \\ = &(n+i+1) \left( \sum_{j = 0}^{i-1} c_{i,j} B_{i,j} +c_{i,i} \right), \end{split} \end{equation}$
其中
$\begin{equation} \begin{split} c_{i,0} &= 1, \\ c_{i,i} &= i\,!, \end{split} \end{equation}$

$\begin{equation} \begin{split} &(n+i+1) c_{i,j} B_{i,j} \\ = &\Big( n-(i+1)+j +1 +2i -j +1 \Big) c_{i,j} B_{i,j} \\ = &c_{i,j} B_{i+1,j} +(2i -j +1) c_{i,j} B_{i+1,j+1}. \end{split} \end{equation}$

由上式可知,
$\begin{equation} \begin{split} c_{i+1,0} = &c_{i,0}, \\ c_{i+1,1} = &c_{i,1} +(2i +1) c_{i,0}, \\ ... \quad & \\ c_{i+1,j} = &c_{i,j} +(2i -j +2) c_{i,j-1}, \\ ... \quad & \\ c_{i+1,i-1} = &c_{i,i-1} +(i+3) c_{i,i-2}, \\ c_{i+1,i} = &c_{i,i} +(i+2) c_{i,i-1}, \\ c_{i+1,i+1} = &(i+1) c_{i,i}. \end{split} \end{equation}$

由于当 $i = 1$ 时, $c_{1,0} = c_{1,1} = 1$ , 因此有
$\begin{equation} \begin{split} c_{i,0} = &1, \\ c_{i,1} = &i^2, \\ c_{i,2} = &\frac{i^2 (i-1)^2}{2}, \\ c_{i,3} = &\frac{i^2 (i-1)^2 (i-2)^2}{3!}, \\ ... \quad & \\ c_{i,j} = &\frac{i^2 (i-1)^2 \cdots (i-j+1)^2}{j\,!}, \\ ... \quad & \\ c_{i,i-2} = &\frac{i^2 (i-1)^2 \cdots 3^2}{(i-2)!} = \frac{i (i-1) i\,!}{2\,!^2}, \\ c_{i,i-1} = &\frac{i^2 (i-1)^2 \cdots 2^2}{(i-1)!} = i\, i\,!, \\ c_{i,i} = & i\,!. \end{split} \end{equation}$

附录

$\begin{equation*} \begin{split} &(n+2) (n+1) \\ = &n (n-1) +4n +2 \\ \\ &(n+3) (n+2) (n+1) \\ = &n (n-1) (n-2) +9n (n-1) +18n +6 \\ \\ &(n+4) (n+3) (n+2) (n+1) \\ = &n (n-1) \cdots (n-3) +16n (n-1) (n-2) +72n (n-1) +96n +24 \\ \\ &(n+5) (n+4) (n+3) (n+2) (n+1) \\ = &n (n-1) \cdots (n-4) +25n (n-1) \cdots (n-3) +200n (n-1) (n-2) +600n (n-1) +600n +120 \end{split} \end{equation*}$

子江 | 随笔

附录

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