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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2項係数の1項間漸化式
\[
C(x+1,y)=\frac{x+1}{x+1-y}C(x,y)
\]
中央2項係数を含む通常型母関数
\[
\sum_{k=0}^{\infty}\frac{1}{k+1}C\left(2k,k\right)z^{k}=\frac{1}{2z}\left\{ 1-\left(1-4z\right)^{\frac{1}{2}}\right\}
\]
2項係数の逆数の差分
\[
C^{-1}(k+j+1,j+1)=\frac{j+1}{j}\left(C^{-1}(k+j,j)-C^{-1}(k+j+1,j)\right)
\]
中央2項係数の通常型母関数
\[
\sum_{k=0}^{\infty}C\left(2k,k\right)z^{k}=\left(1-4z\right)^{-\frac{1}{2}}
\]