Watsonの3F2和公式

まず, Gaussの超幾何定理より,
\begin{align} \frac{(c)_n}{(2c)_n}&=2^{-n}\sum_{0\leq k}\frac{\left(-\frac n2,\frac{1-n}2\right)_k}{k!\left(c+\frac 12\right)_k}\\ &=\sum_{0\leq k}\frac{n!}{2^{n+2k}k!\left(c+\frac 12\right)_k(n-2k)!} \end{align}
であるから,
\begin{align} \sum_{0\leq n}\frac{(a,b,c)_n}{n!\left(\frac{a+b+1}2,2c\right)_n}&=\sum_{0\leq n}\frac{(a,b)_n}{n!\left(\frac{a+b+1}2\right)_n}\sum_{0\leq k}\frac{n!}{2^{n+2k}k!\left(c+\frac 12\right)_k(n-2k)!}\\ &=\sum_{0\leq k}\frac{1}{2^{2k}k!\left(c+\frac 12\right)_k}\sum_{0\leq n}\frac{(a,b)_n}{2^n\left(\frac{a+b+1}2\right)_n(n-2k)!}\\ \end{align}
ここで, Gaussによる公式
\begin{align} \sum_{0\leq n}\frac{(a,b)_n}{2^nn!\left(\frac{a+b+1}2\right)_n}&=\frac{\Gamma\left(\frac 12\right)\Gamma\left(\frac{a+b+1}2\right)}{\Gamma\left(\frac{a+1}2\right)\Gamma\left(\frac{b+1}2\right)} \end{align}
を用いて,

\begin{align} \sum_{0\leq n}\frac{(a,b)_n}{2^n\left(\frac{a+b+1}2\right)_n(n-2k)!}&=\frac{(a,b)_{2k}}{2^{2k}\left(\frac{a+b+1}2\right)_{2k}}\sum_{0\leq n}\frac{(a+2k,b+2k)_n}{2^nn!\left(\frac{a+b+1}2+2k\right)_n}\\ &=\frac{(a,b)_{2k}}{2^{2k}\left(\frac{a+b+1}2\right)_{2k}}\frac{\Gamma\left(\frac 12\right)\Gamma\left(\frac{a+b+1}2+2k\right)}{\Gamma\left(\frac{a+1}2+k\right)\Gamma\left(\frac{b+1}2+k\right)}\\ &=\frac{\Gamma\left(\frac 12\right)\Gamma\left(\frac{a+b+1}2\right)}{\Gamma\left(\frac{a+1}2\right)\Gamma\left(\frac{b+1}2\right)}\frac{(a,b)_{2k}}{2^{2k}\left(\frac{a+1}2,\frac{b+1}2\right)_k}\\ &=\frac{\Gamma\left(\frac 12\right)\Gamma\left(\frac{a+b+1}2\right)}{\Gamma\left(\frac{a+1}2\right)\Gamma\left(\frac{b+1}2\right)}2^{2k}\left(\frac a2,\frac b2\right)_{k} \end{align}
だから,
\begin{align} \sum_{0\leq k}\frac{1}{2^{2k}k!\left(c+\frac 12\right)_k}\sum_{0\leq n}\frac{(a,b)_n}{2^n\left(\frac{a+b+1}2\right)_n(n-2k)!}&=\frac{\Gamma\left(\frac 12\right)\Gamma\left(\frac{a+b+1}2\right)}{\Gamma\left(\frac{a+1}2\right)\Gamma\left(\frac{b+1}2\right)}\sum_{0\leq k}\frac{\left(\frac a2,\frac b2\right)_{k}}{k!\left(c+\frac 12\right)_k} \end{align}
ここで, Gaussの超幾何定理から
\begin{align} \sum_{0\leq k}\frac{\left(\frac a2,\frac b2\right)_{k}}{k!\left(c+\frac 12\right)_k}&=\frac{\Gamma\left(c+\frac 12\right)\Gamma\left(c+\frac{1-a-b}2\right)}{\Gamma\left(c+\frac{1-a}2\right)\Gamma\left(c+\frac{1-b}2\right)} \end{align}
であるから定理を得る.