3F2と2F1の有限和の恒等式

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$\sum_{k = 0}^{n} (- 4)^{k} (\binom{n}{k}) \frac{(a, b)_{k}}{(a + b)_{2 k}} = \sum_{k = 0}^{2 n} (- 2)^{k} (\binom{2 n}{k}) \frac{(a)_{k}}{(a + b)_{k}}$

$(a_{1}, \dots, a_{j})_{n} : = (a_{1})_{n} \dots (a_{j})_{n}$ と略記します。

超幾何級数を使うとこのようになります。
$_{3} F_{2} [\begin{matrix} a, b, - n \\ \frac{a + b}{2}, \frac{a + b + 1}{2} \end{matrix}; 1] =_{2} F_{1} [\begin{matrix} a, - 2 n \\ a + b \end{matrix}; 2]$

$\begin{array}{rcl} \frac{B (a + k, b + l)}{B (a, b)} & = & \frac{Γ (a + b)}{Γ (a) Γ (b)} \frac{Γ (a + k) Γ (b + l)}{Γ (a + b + k + l)} \\ = & \frac{(a)_{k} (b)_{l}}{(a + b)_{k + l}} \end{array}$
$\begin{array}{rcl} (左辺) & = & \sum_{k = 0}^{n} (\binom{n}{k}) \frac{B (a + k, b + k)}{B (a, b)} (- 4)^{k} \\ = & \frac{1}{B (a, b)} \int_{0}^{1} t^{a - 1} (1 - t)^{b - 1} [1 - 4 t (1 - t)]^{n} d t \\ = & \frac{1}{B (a, b)} \int_{0}^{1} t^{a - 1} (1 - t)^{b - 1} {(1 - 4 t + 4 t^{2})}^{n} d t \\ = & \frac{1}{B (a, b)} \int_{0}^{1} t^{a - 1} (1 - t)^{b - 1} (1 - 2 t)^{2 n} d t \\ = & \sum_{k = 0}^{2 n} (- 2)^{k} (\binom{2 n}{k}) \frac{B (a + k, b)}{B (a, b)} \\ = & (右辺) \end{array}$

$\begin{aligned} \sum_{n = 0}^{\infty} \frac{(a, b)_{n}}{(a + b)_{2 n}} {(- \frac{4 x^{2}}{1 - x^{2}})}^{n} & = \frac{1}{2} \sum_{n = 0}^{\infty} \frac{(a)_{n}}{(a + b)_{n}} [(1 + x) {(- \frac{2 x}{1 - x})}^{n} + (1 - x) {(\frac{2 x}{1 + x})}^{n}] \\ = \frac{1 \pm x}{2} \sum_{n = 0}^{\infty} \frac{(a)_{n} + (b)_{n}}{(a + b)_{n}} {(\mp \frac{2 x}{1 \mp x})}^{n} \end{aligned}$

$\begin{aligned} ∵ \\ f (x) = \sum_{n = 0}^{\infty} a_{n} x^{n} \\ ⟺ \\ \frac{1}{1 - x} f (\frac{x}{1 - x}) = \sum_{n = 0}^{\infty} [\sum_{k = 0}^{n} (\binom{n}{k}) a_{k}] x^{n} \end{aligned}$

これは超幾何関数を使うと
$\begin{aligned} _{3} F_{2} [\begin{array}{c} a, b, 1 \\ \frac{a + b}{2}, \frac{a + b + 1}{2} \end{array}; - \frac{x^{2}}{1 - x^{2}}] & = \frac{1}{2} [(1 + x)_{2} F_{1} [\begin{array}{c} a, 1 \\ a + b \end{array}; - \frac{2 x}{1 - x}] + (1 - x)_{2} F_{1} [\begin{array}{c} a, 1 \\ a + b \end{array}; \frac{2 x}{1 + x}]] \\ = \frac{1 \pm x}{2} (_{2} F_{1} [\begin{array}{c} a, 1 \\ a + b \end{array}; \mp \frac{2 x}{1 \mp x}] +_{2} F_{1} [\begin{array}{c} b, 1 \\ a + b \end{array}; \mp \frac{2 x}{1 \mp x}]) \end{aligned}$
となります。

$_{3} F_{2} [\begin{matrix} - n, b, c \\ \frac{- n + b}{2}, \frac{- n + b + 1}{2} \end{matrix}; 1] =_{2} F_{1} [\begin{matrix} - n, 2 c \\ - n + b \end{matrix}; 2]$