Watsonの3F2和公式

まず, Gaussの超幾何定理より,
$$\begin{array}{rl}\frac{(c{)}_n}{(2c{)}_n}& =2^{-n}\sum _{0\le k}\frac{{(-\frac{n}{2},\frac{1-n}{2})}_k}{k!{(c+\frac{1}{2})}_k}\\ & =\sum _{0\le k}\frac{n!}{2^{n+2k}k!{(c+\frac{1}{2})}_k(n-2k)!}\end{array}$$
であるから,
$$\begin{array}{rl}\sum _{0\le n}\frac{(a,b,c{)}_n}{n!{(\frac{a+b+1}{2},2c)}_n}& =\sum _{0\le n}\frac{(a,b{)}_n}{n!{\left(\frac{a+b+1}{2}\right)}_n}\sum _{0\le k}\frac{n!}{2^{n+2k}k!{(c+\frac{1}{2})}_k(n-2k)!}\\ & =\sum _{0\le k}\frac{1}{2^{2k}k!{(c+\frac{1}{2})}_k}\sum _{0\le n}\frac{(a,b{)}_n}{2^n{\left(\frac{a+b+1}{2}\right)}_n(n-2k)!}\end{array}$$
ここで, Gaussによる公式
$$\begin{array}{rl}\sum _{0\le n}\frac{(a,b{)}_n}{2^{n}n!{\left(\frac{a+b+1}{2}\right)}_n}& =\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }\left(\frac{a+b+1}{2}\right)}{\mathrm{\Gamma }\left(\frac{a+1}{2}\right)\mathrm{\Gamma }\left(\frac{b+1}{2}\right)}\end{array}$$
を用いて,

$$\begin{array}{rl}\sum _{0\le n}\frac{(a,b{)}_n}{2^n{\left(\frac{a+b+1}{2}\right)}_n(n-2k)!}& =\frac{(a,b{)}_{2k}}{2^{2k}{\left(\frac{a+b+1}{2}\right)}_{2k}}\sum _{0\le n}\frac{(a+2k,b+2k{)}_n}{2^{n}n!{(\frac{a+b+1}{2}+2k)}_n}\\ & =\frac{(a,b{)}_{2k}}{2^{2k}{\left(\frac{a+b+1}{2}\right)}_{2k}}\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }(\frac{a+b+1}{2}+2k)}{\mathrm{\Gamma }(\frac{a+1}{2}+k)\mathrm{\Gamma }(\frac{b+1}{2}+k)}\\ & =\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }\left(\frac{a+b+1}{2}\right)}{\mathrm{\Gamma }\left(\frac{a+1}{2}\right)\mathrm{\Gamma }\left(\frac{b+1}{2}\right)}\frac{(a,b{)}_{2k}}{2^{2k}{(\frac{a+1}{2},\frac{b+1}{2})}_k}\\ & =\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }\left(\frac{a+b+1}{2}\right)}{\mathrm{\Gamma }\left(\frac{a+1}{2}\right)\mathrm{\Gamma }\left(\frac{b+1}{2}\right)}{2}^{2k}{(\frac{a}{2},\frac{b}{2})}_k\end{array}$$
だから,
$$\begin{array}{rl}\sum _{0\le k}\frac{1}{2^{2k}k!{(c+\frac{1}{2})}_k}\sum _{0\le n}\frac{(a,b{)}_n}{2^n{\left(\frac{a+b+1}{2}\right)}_n(n-2k)!}& =\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }\left(\frac{a+b+1}{2}\right)}{\mathrm{\Gamma }\left(\frac{a+1}{2}\right)\mathrm{\Gamma }\left(\frac{b+1}{2}\right)}\sum _{0\le k}\frac{{(\frac{a}{2},\frac{b}{2})}_k}{k!{(c+\frac{1}{2})}_k}\end{array}$$
ここで, Gaussの超幾何定理から
$$\begin{array}{rl}\sum _{0\le k}\frac{{(\frac{a}{2},\frac{b}{2})}_k}{k!{(c+\frac{1}{2})}_k}& =\frac{\mathrm{\Gamma }(c+\frac{1}{2})\mathrm{\Gamma }(c+\frac{1-a-b}{2})}{\mathrm{\Gamma }(c+\frac{1-a}{2})\mathrm{\Gamma }(c+\frac{1-b}{2})}\end{array}$$
であるから定理を得る.