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/ |
SNSボタン |
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項係数の特殊な積
\[
C(x,t)C(t,y)=C(x,y)C(x-y,x-t)
\]
ファンデルモンドの畳み込み定理と第1引数の畳み込み
\[
\sum_{j=0}^{k}C(x,j)C(y,k-j)=C(x+y,k)
\]
2項係数の母関数
\[
\sum_{k=0}^{\infty}C(x+k,k)t^{k}=(1-t)^{-(x+1)}
\]