1次式の総乗と階乗

by nomura · 2021年2月3日

1次式の総乗と階乗
\(a,b\in\mathbb{Z}\)とする。
\[ \prod_{k=a}^{b}\left(kn+r\right)=n^{b-a+1}\frac{\left(b+\frac{r}{n}\right)!}{\Gamma\left(a+\frac{r}{n}\right)} \]

(0)

\begin{align*} \prod_{k=a}^{b}\left(kn+r\right) & =n^{b-a+1}\prod_{k=a}^{b}\left(k+\frac{r}{n}\right)\\ & =n^{b-a+1}\prod_{k=a}^{b}\frac{\left(k+\frac{r}{n}\right)!}{\left(k+\frac{r}{n}-1\right)!}\\ & =n^{b-a+1}\frac{\left(b+\frac{r}{n}\right)!}{\Gamma\left(a+\frac{r}{n}\right)} \end{align*}

(0)-2

\begin{align*} \prod_{k=a}^{b}\left(kn+r\right) & =\prod_{k=a}^{-1}\left(kn+r\right)\prod_{k=0}^{b}\left(kn+r\right)\\ & =\prod_{k=0}^{a-1}\left(kn+r\right)^{-1}\prod_{k=0}^{b}\left(kn+r\right)\\ & =\left\{ n^{a-1}r\frac{\left(a-1+\frac{r}{n}\right)!}{\frac{r}{n}!}\right\} ^{-1}n^{b}r\frac{\left(b+\frac{r}{n}\right)!}{\frac{r}{n}!}\\ & =n^{b-a+1}\frac{\left(b+\frac{r}{n}\right)!}{\left(a+\frac{r}{n}-1\right)!}\\ & =n^{b-a+1}\frac{\left(b+\frac{r}{n}\right)!}{\Gamma\left(a+\frac{r}{n}\right)} \end{align*}