パスカルの法則の応用
パスカルの法則の応用
\(n\in\mathbb{N}_{0}\)とする。
\(n\in\mathbb{N}_{0}\)とする。
(1)
\[ C\left(x+n,y+n\right)=C\left(x,y+n\right)+\sum_{k=0}^{n-1}C\left(x+k,y+n-1\right) \](2)
\[ C\left(x+n,y+n\right)=C\left(x,y\right)+\sum_{k=0}^{n-1}C\left(x+k,y+k+1\right) \](3)
\[ C\left(x+n,y+n\right)=\left(-1\right)^{n}\left\{ C\left(x+n,y\right)-\sum_{k=0}^{n-1}\left(-1\right)^{k}C\left(x+n+1,y+1+k\right)\right\} \](1)
\begin{align*} C\left(x+n,y+n\right) & =C\left(x+n-1,y+n\right)+C\left(x+n-1,y+n-1\right)\\ & =C\left(x,y+n\right)+\sum_{k=0}^{n-1}\left\{ C\left(x+n-k,y+n\right)-C\left(x+n-\left(k+1\right),y+n\right)\right\} \\ & =C\left(x,y+n\right)+\sum_{k=0}^{n-1}C\left(x+n-\left(k+1\right),y+n-1\right)\\ & =C\left(x,y+n\right)+\sum_{k=1}^{n}C\left(x+n-k,y+n-1\right)\\ & =C\left(x,y+n\right)+\sum_{k=0}^{n-1}C\left(x+k,y+n-1\right) \end{align*}(2)
\begin{align*} C\left(x+n,y+n\right) & =C\left(x+n-1,y+n\right)+C\left(x+n-1,y+n-1\right)\\ & =C\left(x,y\right)+\sum_{k=0}^{n-1}\left\{ C\left(x+n-k,y+n-k\right)-C\left(x+n-\left(k+1\right),y+n-\left(k+1\right)\right)\right\} \\ & =C\left(x,y\right)+\sum_{k=0}^{n-1}C\left(x+n-\left(k+1\right),y+n-k\right)\\ & =C\left(x,y\right)+\sum_{k=0}^{n-1}C\left(x+k,y+k+1\right) \end{align*}(3)
\begin{align*} C\left(x+n,y+n\right) & =-C\left(x+n,y+n-1\right)+C\left(x+n+1,y+n\right)\\ & =\left(-1\right)^{n}C\left(x+n,y\right)-\sum_{k=0}^{n-1}\left\{ \left(-1\right)^{k+1}C\left(x+n,y+n-k\right)-\left(-1\right)^{k}C\left(x+n,y+n-1-k\right)\right\} \\ & =\left(-1\right)^{n}C\left(x+n,y\right)-\sum_{k=0}^{n-1}\left(-1\right)^{k+1}\left\{ C\left(x+n,y+n-k\right)+C\left(x+n,y+n-1-k\right)\right\} \\ & =\left(-1\right)^{n}C\left(x+n,y\right)-\sum_{k=0}^{n-1}\left(-1\right)^{k+1}\left\{ C\left(x+n+1,y+n-k\right)\right\} \\ & =\left(-1\right)^{n}C\left(x+n,y\right)+\sum_{k=0}^{n-1}\left(-1\right)^{k}C\left(x+n+1,y+n-k\right)\\ & =\left(-1\right)^{n}C\left(x+n,y\right)+\sum_{k=0}^{n-1}\left(-1\right)^{n-1-k}C\left(x+n+1,y+1+k\right)\\ & =\left(-1\right)^{n}\left\{ C\left(x+n,y\right)-\sum_{k=0}^{n-1}\left(-1\right)^{k}C\left(x+n+1,y+1+k\right)\right\} \end{align*}
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2項変換と交代2項変換の母関数
\[
\sum_{k=0}^{\infty}b_{k}x^{k}=\frac{1}{1-x}\sum_{k=0}^{\infty}a_{k}\left(\frac{x}{1-x}\right)^{k}
\]
負の整数の2項係数
\[
C\left(-m,-n\right)=\left(-1\right)^{m-n}C\left(n-1,m-1\right)
\]
2項係数の逆数の差分
\[
C^{-1}(k+j+1,j+1)=\frac{j+1}{j}\left(C^{-1}(k+j,j)-C^{-1}(k+j+1,j)\right)
\]
2項係数の半分までの総和
\[
\sum_{k=0}^{n-1}C\left(2n-1,k\right)=2^{2n-2}
\]