2項係数の1項間漸化式
2項係数は以下の1項間漸化式を満たす。
(1)
\[ C(x+1,y)=\frac{x+1}{x+1-y}C(x,y) \](2)
\[ C(x-1,y)=\frac{x-y}{x}C(x,y) \](3)
\[ C(x,y+1)=\frac{x-y}{y+1}C(x,y) \](4)
\[ C(x,y-1)=\frac{y}{x-y+1}C(x,y) \](5)
\[ C(x+1,y+1)=\frac{x+1}{y+1}C(x,y) \](6)
\[ C(x+1,y-1)=\frac{(x+1)y}{(x-y+1)(x-y+2)}C(x,y) \](7)
\[ C(x-1,y+1)=\frac{(x-y-1)(x-y)}{x(y+1)}C(x,y) \](8)
\[ C(x-1,y-1)=\frac{y}{x}C(x,y) \](1)
\begin{align*} C(x+1,y) & =\frac{(x+1)!}{y!(x+1-y)!}\\ & =\frac{x+1}{x+1-y}\frac{x!}{y!(x-y)!}\\ & =\frac{x+1}{x+1-y}C(x,y) \end{align*}(2)
\begin{align*} C(x-1,y) & =\frac{(x-1)!}{y!(x-1-y)!}\\ & =\frac{x-y}{x}\frac{x!}{y!(x-y)!}\\ & =\frac{x-y}{x}C(x,y) \end{align*}(3)
\begin{align*} C(x,y+1) & =\frac{x!}{(y+1)!(x-y-1)!}\\ & =\frac{x-y}{y+1}\frac{x!}{y!(x-y)!}\\ & =\frac{x-y}{y+1}C(x,y) \end{align*}(4)
\begin{align*} C(x,y-1) & =\frac{x!}{(y-1)!(x-y+1)!}\\ & =\frac{y}{x-y+1}\frac{x!}{y!(x-y)!}\\ & =\frac{y}{x-y+1}C(x,y) \end{align*}(5)
\begin{align*} C(x+1,y+1) & =\frac{(x+1)!}{(y+1)!(x-y)!}\\ & =\frac{x+1}{y+1}\frac{x!}{y!(x-y)!}\\ & =\frac{x+1}{y+1}C(x,y) \end{align*}(6)
\begin{align*} C(x+1,y-1) & =\frac{(x+1)!}{(y-1)!(x-y+2)!}\\ & =\frac{(x+1)y}{(x-y+1)(x-y+2)}\frac{x!}{y!(x-y)!}\\ & =\frac{(x+1)y}{(x-y+1)(x-y+2)}C(x,y) \end{align*}(7)
\begin{align*} C(x-1,y+1) & =\frac{(x-1)!}{(y+1)!(x-y-2)!}\\ & =\frac{(x-y-1)(x-y)}{x(y+1)}\frac{x!}{y!(x-y)!}\\ & =\frac{(x-y-1)(x-y)}{x(y+1)}C(x,y) \end{align*}(8)
\begin{align*} C(x-1,y-1) & =\frac{(x-1)!}{(y-1)!(x-y)!}\\ & =\frac{y}{x}\frac{x!}{y!(x-y)!}\\ & =\frac{y}{x}C(x,y) \end{align*}ページ情報
| タイトル | 2項係数の1項間漸化式 |
| URL | https://www.nomuramath.com/p8gtr0e4/ |
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2項係数が0になるとき
\[
\forall m,n\in\mathbb{Z},\left(0\leq m<n\right)\lor\left(n<0\leq m\right)\lor\left(m<n<0\right)\Leftrightarrow C\left(m,n\right)=0
\]
2項係数の半分までの総和
\[
\sum_{k=0}^{n-1}C\left(2n-1,k\right)=2^{2n-2}
\]
2項変換と交代2項変換の逆変換
\[
a_{n}=\sum_{k=0}^{n}(-1)^{n-k}C(n,k)b_{k}
\]
2項係数の2乗和
\[
\sum_{j=0}^{m}C^{2}(m,j)=C(2m,m)
\]