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In the solution to the Newton-Pepys problem here, there's an identity applied on the third line to relate a binomial sum to the hypergeometric function 2F1. What is the general statement of the identity?

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up vote 2 down vote accepted

$$P_n = \sum _{x=n}^{6n} \binom{6n}{x}\left(\frac{1}{6}\right)^{x}\left(\frac{5}{6}\right)^{6n-x}\\ =\frac{5^{5n}}{6^{6n}}\binom{6n}{n}\sum_{x=n}^{6n}\frac{(5n)!n!}{x!(6n-x)!}5^{n-x}\\ =\frac{5^{5n}}{6^{6n}}\binom{6n}{n}\sum_{k=0}^{5n}\frac{(5n)!n!}{(k+n)!(5n-k)!}\frac{1}{5^k}\\ =\frac{5^{5n}}{6^{6n}}\binom{6n}{n}\sum_{k=0}^{5n}\frac{(5n-k+1)\cdots(5n)}{(n+1)\cdots(n+k)}\frac{1}{5^k}\\ =\frac{5^{5n}}{6^{6n}}\binom{6n}{n}\sum_{k=0}^{5n}\frac{(-5n)_k}{(n+1)_k}\left(-\frac{1}{5}\right)^k\frac{k!}{k!}\\ =\frac{5^{5n}}{6^{6n}}\binom{6n}{n}\sum_{k=0}^{5n}\frac{(-5n)_k(1)_k}{(n+1)_k}\left(-\frac{1}{5}\right)^k\frac{1}{k!}\\ =\frac{5^{5n}}{6^{6n}}\binom{6n}{n}{_2F_1}(1;-5n;n+1;-\frac{1}{5}).$$ I used this symbol as in Wikipedia's definition of hypergeometric series.

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