2項係数の半分までの総和
2項係数の半分までの総和
(1)偶数の場合で半分以下
\[ \sum_{k=0}^{n-1}C\left(2n,k\right)=2^{2n-1}-\frac{1}{2}C\left(2n,n\right) \](2)偶数の場合で半分以上
\[ \sum_{k=0}^{n}C\left(2n,k\right)=2^{2n-1}+C\left(2n,n\right) \](3)奇数の場合で丁度半分
\[ \sum_{k=0}^{n-1}C\left(2n-1,k\right)=2^{2n-2} \](1)
\begin{align*} \sum_{k=0}^{n-1}C\left(2n,k\right) & =\frac{1}{2}\left(\sum_{k=0}^{n-1}C\left(2n,k\right)+\sum_{k=0}^{n-1}C\left(2n,2n-k\right)\right)\\ & =\frac{1}{2}\left(\sum_{k=0}^{n-1}C\left(2n,k\right)+\sum_{k=0}^{n-1}C\left(2n,2n-\left(n-1-k\right)\right)\right)\\ & =\frac{1}{2}\left(\sum_{k=0}^{n-1}C\left(2n,k\right)+\sum_{k=0}^{n-1}C\left(2n,n+1+k\right)\right)\\ & =\frac{1}{2}\left(\sum_{k=0}^{n-1}C\left(2n,k\right)+\sum_{k=n+1}^{2n}C\left(2n,k\right)\right)\\ & =\frac{1}{2}\left(\sum_{k=0}^{2n}C\left(2n,k\right)-C\left(2n,n\right)\right)\\ & =2^{2n-1}-\frac{1}{2}C\left(2n,n\right) \end{align*}(2)
\begin{align*} \sum_{k=0}^{n}C\left(2n,k\right) & =\sum_{k=0}^{n-1}C\left(2n,k\right)+C\left(2n,n\right)\\ & =2^{2n-1}-\frac{1}{2}C\left(2n,n\right)+C\left(2n,n\right)\\ & =2^{2n-1}+C\left(2n,n\right) \end{align*}(3)
\begin{align*} \sum_{k=0}^{n-1}C\left(2n-1,k\right) & =\frac{1}{2}\left(\sum_{k=0}^{n-1}C\left(2n-1,k\right)+\sum_{k=0}^{n-1}C\left(2n-1,2n-1-k\right)\right)\\ & =\frac{1}{2}\left(\sum_{k=0}^{n-1}C\left(2n-1,k\right)+\sum_{k=0}^{n-1}C\left(2n-1,2n-1-\left(n-1-k\right)\right)\right)\\ & =\frac{1}{2}\left(\sum_{k=0}^{n-1}C\left(2n-1,k\right)+\sum_{k=0}^{n-1}C\left(2n-1,n+k\right)\right)\\ & =\frac{1}{2}\left(\sum_{k=0}^{n-1}C\left(2n-1,k\right)+\sum_{k=n}^{2n-1}C\left(2n-1,k\right)\right)\\ & =\frac{1}{2}\left(\sum_{k=0}^{2n-1}C\left(2n-1,k\right)\right)\\ & =2^{2n-2} \end{align*}ページ情報
タイトル | 2項係数の半分までの総和 |
URL | https://www.nomuramath.com/k1y1011c/ |
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パスカルの法則の応用
\[
C\left(x+n,y+n\right)=C\left(x,y+n\right)+\sum_{k=0}^{n-1}C\left(x+k,y+n-1\right)
\]
パスカルの法則の一般形
\[
C\left(x+n,y+n\right)=\sum_{k=0}^{n}C\left(n,k\right)C\left(x,y+k\right)
\]
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
\[
C^{-1}(k+j+1,j+1)=\frac{j+1}{j}\left(C^{-1}(k+j,j)-C^{-1}(k+j+1,j)\right)
\]