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項係数の飛び飛びの総和
\[
\sum_{k=-\infty}^{\infty}C\left(mn,mk+l\right)=\frac{1}{m}\sum_{j=0}^{m-1}\left(1+\omega_{m}^{j}\right)^{mn}\left(\omega_{m}^{j}\right)^{-l}
\]
ディクソンの等式
\[
\sum_{k=-a}^{a}(-1)^{k}C(a+b,a+k)C(b+c,b+k)C(c+a,c+k)=\frac{(a+b+c)!}{a!b!c!}
\]
2項係数の母関数
\[
\sum_{k=0}^{\infty}C(x+k,k)t^{k}=(1-t)^{-(x+1)}
\]