2項係数を含む総和
2項係数を含む総和
(1)
\[ \sum_{k=0}^{n}\frac{\left(-1\right)^{k}C\left(n,k\right)}{m+k}=\frac{1}{mC\left(m+n,m\right)} \](2)
\[ \sum_{k=0}^{n}\frac{\left(-1\right)^{k}C\left(n,k\right)}{m-k}=\frac{\left(-1\right)^{n}}{\left(m-n\right)C\left(m,n\right)} \](1)
\begin{align*} \sum_{k=0}^{n}\frac{\left(-1\right)^{k}C\left(n,k\right)}{m+k} & =\int_{0}^{1}\sum_{k=0}^{n}\left(-1\right)^{k}x^{m+k-1}C\left(n,k\right)dx\\ & =\left(-1\right)^{-n}\int_{0}^{1}x^{m-1}\sum_{k=0}^{n}\left(-1\right)^{n-k}x^{k}C\left(n,k\right)dx\\ & =\left(-1\right)^{-n}\int_{0}^{1}x^{m-1}\left(x-1\right)^{n}dx\\ & =\int_{0}^{1}x^{m-1}\left(1-x\right)^{n}dx\\ & =B\left(m,n+1\right)\\ & =\frac{\Gamma\left(m\right)\Gamma\left(n+1\right)}{\Gamma\left(m+n+1\right)}\\ & =\frac{\Gamma\left(m+1\right)\Gamma\left(n+1\right)}{m\Gamma\left(m+n+1\right)}\\ & =\frac{1}{mC\left(m+n,m\right)} \end{align*}(2)
\begin{align*} \sum_{k=0}^{n}\frac{\left(-1\right)^{k}C\left(n,k\right)}{m-k} & =\int_{0}^{1}\sum_{k=0}^{n}\left(-1\right)^{k}x^{m-k-1}C\left(n,k\right)dx\\ & =\int_{0}^{1}x^{m-n-1}\sum_{k=0}^{n}\left(-1\right)^{k}x^{n-k}C\left(n,k\right)dx\\ & =\int_{0}^{1}x^{m-n-1}\left(x-1\right)^{n}dx\\ & =\left(-1\right)^{n}\int_{0}^{1}x^{m-n-1}\left(1-x\right)^{n}dx\\ & =\left(-1\right)^{n}B\left(m-n,n+1\right)\\ & =\left(-1\right)^{n}\frac{\Gamma\left(m-n\right)\Gamma\left(n+1\right)}{\Gamma\left(m+1\right)}\\ & =\left(-1\right)^{n}\frac{\Gamma\left(m-n+1\right)\Gamma\left(n+1\right)}{\left(m-n\right)\Gamma\left(m+1\right)}\\ & =\frac{\left(-1\right)^{n}}{\left(m-n\right)C\left(m,n\right)} \end{align*}ページ情報
タイトル | 2項係数を含む総和 |
URL | https://www.nomuramath.com/qul675uc/ |
SNSボタン |
2項係数の母関数
\[
\sum_{k=0}^{\infty}C(x+k,k)t^{k}=(1-t)^{-(x+1)}
\]
2項係数の逆数の差分
\[
C^{-1}(k+j+1,j+1)=\frac{j+1}{j}\left(C^{-1}(k+j,j)-C^{-1}(k+j+1,j)\right)
\]
飛び飛びの2項定理
\[
\sum_{k=0}^{\infty}C\left(n,2k\right)a^{2k}b^{n-2k}=\frac{1}{2}\left\{ \left(a+b\right)^{n}+\left(-a+b\right)^{n}\right\}
\]
パスカルの法則
\[
C(x+1,y+1)=C(x,y+1)+C(x,y)
\]