1+(1/(16))+(5^2 /(16^2 .2!))+((5^2 .9^2 )/(16^3 .3!))+((5^2 .9^2 .13^2 )/(16^4 .4!))+...=((√π)/(Γ^2 ((3/4))))=F_1 ((1/4),(1/4),1;1) Prove The above relation Where F_1 (Φ,ϕ,γ;μ)=Σ_(n≥0) ^∞ (((Φ)_n (ϕ)_n )/(n!(γ)_n ))μ^n (ζ)


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Question Number 128122 by Dwaipayan Shikari last updated on 04/Jan/21

$1 + \frac{1}{16} + \frac{5^{2}}{16^{2} . 2!} + \frac{5^{2} . 9^{2}}{16^{3} . 3!} + \frac{5^{2} . 9^{2} . 13^{2}}{16^{4} . 4!} + . . . = \frac{\sqrt{π}}{Γ^{2} (\frac{3}{4})} = F_{1} (\frac{1}{4}, \frac{1}{4}, 1; 1)$ $P r o v e T h e a b o v e r e l a t i o n$ $W h e r e$ $F_{1} (Φ, φ, γ; μ) = \underset{n ⩾ 0}{\sum^{\infty}} \frac{{(Φ)}_{n} {(φ)}_{n}}{n! {(γ)}_{n}} μ^{n}$ ${(ζ)}_{n} = ζ (ζ + 1) (ζ + 2) . . . (ζ + n - 1)$
	Answered by mindispower last updated on 05/Jan/21

	$= 1 + \sum_{n ⩾ 1} \frac{\underset{k = 0}{\prod^{n - 1}} {(4 k + 1)}^{2}}{16^{n} {(n!)}^{2}}$ $= 1 + \sum_{n ⩾ 1} \frac{4^{2 n} . \underset{k = 0}{\prod^{n - 1}} (k + \frac{1}{4}) . Π (k + \frac{1}{4})}{. 16^{n} . n!} . \frac{1}{n!}$ $= 1 + \sum_{n ⩾ 1} \frac{{(\frac{1}{4})}_{n} . {(\frac{1}{4})}_{n}}{{(1)}_{n}} . \frac{1^{n}}{n!} =_{2} F_{1} (\frac{1}{4}, \frac{1}{4}; 1; [1])$ $t h a n u s e r e l a t i o n_{2} F_{1} (a, b, c; 1) w i t h e β,$ $Γ$
	Commented by Dwaipayan Shikari last updated on 05/Jan/21

	$G r e a t s i r!$ $t h e n_{2} F_{1} (a, b, c; 1) = \frac{Γ (c) Γ (c - a - b)}{Γ (c - a) Γ (c - b)}$ $_{2} F_{1} (\frac{1}{4}, \frac{1}{4}; 1; 1) = \frac{Γ (\frac{1}{2})}{Γ^{2} (\frac{3}{4})} = \frac{\sqrt{π}}{Γ^{2} (\frac{3}{4})}$ $I s i t s i r ? ?$
	Commented by mindispower last updated on 06/Jan/21

	$y e s s i r g r e a t$
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