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/ |
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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項係数の第1引数と第2引数同士の総和
\[
\sum_{j=0}^{k-a}\left(-1\right)^{j}C\left(k,j+a\right)C\left(j+b,c\right)=\begin{cases}
\left(-1\right)^{k-a}C\left(b-a,c-k\right) & a-b+c\leq k\\
0 & k<a-b+c
\end{cases}
\]
2項係数の1項間漸化式
\[
C(x+1,y)=\frac{x+1}{x+1-y}C(x,y)
\]
ディクソンの等式
\[
\sum_{k=-a}^{a}(-1)^{k}C(a+b,a+k)C(b+c,b+k)C(c+a,c+k)=\frac{(a+b+c)!}{a!b!c!}
\]

