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Question Number 127725 by mnjuly1970 last updated on 01/Jan/21

$. . . a d v a n c e d m a t h e m a t i c s . . .$ $p r o v e t h a t ::$ $\underset{n = 0}{\sum^{\infty}} \frac{1}{2^{n} (\begin{matrix} 3 n \\ n \end{matrix})} \overset{? ? ?}{=} \frac{3}{125} (\frac{11 π}{6} - 2 l o g (2) + 45)$
	Answered by Ar Brandon last updated on 01/Jan/21
	$1+((2!)/(2∙3!))+((4!2!)/(2^2 ∙6!))+((6!3!)/(2^3 9!))+∙∙∙(((2n)!n!)/(2^n (3n)!))+∙∙∙ S_n =Σ_(n=0) ^∞ ((Γ(2n+1)Γ(n+1))/(2^n Γ(3n+1)))=Σ_(n=0) ^∞ ((nΓ(2n+1)Γ(n))/(2^n Γ(3n+1))) =Σ_(n=0) ^∞ (n/2^n )β(2n+1, n)=Σ_(n=0) ^∞ (n/2^n )∫_0 ^1 x^(2n) (1−x)^(n−1) dx =∫_0 ^1 {Σ_(n=0) ^∞ [(n/2^n )x^(2n) (1−x)^(n−1) ]}dx=∫_0 ^1 {(1/(1−x))∙Σ_(n=0) ^∞ n(((x^2 (1−x))/2))^n }dx S(t)=Σ_(k=0) ^∞ t^k =(1/(1−t)) ⇒S′(t)=Σ_(k=0) ^∞ kt^(k−1) =(1/((1−t)^2 )) tS′(t)=Σ_(k=0) ^∞ kt^k =(t/((1−t)^2 )) Σ_(n=0) ^∞ n(((x^2 (1−x))/2))^n =(((x^2 (1−x))/2)/((1−((x^2 (1−x))/2))^2 ))=((x^2 (1−x))/2)∙(4/((2−x^2 +x^3 )^2 )) =((2x^2 (1−x))/((x^3 −x^2 +2)^2 )) S_n =∫_0 ^1 {(1/(1−x))∙((2x^2 (1−x))/((x^3 −x^2 +2)^2 ))}dx=∫_0 ^1 ((2x^2 dx)/((x^3 −x^2 +2)^2 )) =∫_0 ^1 ((2x^2 )/(((x+1)(x^2 −2x+2))^2 )) f(x)=2((x/((x+1)(x^2 −2x+2))))^2 =(2/(25))(((x+2)/(x^2 −2x+2))−(1/(x+1)))^2 =(2/(25))(((x^2 +4x+4)/((x^2 −2x+2)^2 ))−((2(x+2))/((x+1)(x^2 −2x+2)))+(1/((x+1)^2 ))) =(2/(25))((1/(x^2 −2x+2))+((6x+2)/((x^2 −2x+2)^2 ))−(2/(5(x+1)))+((2(x−1))/(x^2 −2x+2))+(1/((x+1)^2 ))) 6x+2=3(2x−2)+8 ∫_0 ^1 f(x)dx=(2/(25))[tan^(−1) (x−1)−(3/(x^2 −2x+2))+8∫(dx/(((x−1)^2 +1)^2 )) [−((2ln(x+1))/5)−ln(x^2 −2x+2)−(1/(x+1))]_0 ^1 S_n =(2/(25))(−3−((2ln2)/5)−(1/2))−(2/(25))(−(π/4)−(3/2)−ln2−1)−((2π)/(25)) =−(2/(25))+((6ln2)/(125))−((6π)/(100)) Typos$
	$1 + \frac{2!}{2 \cdot 3!} + \frac{4! 2!}{2^{2} \cdot 6!} + \frac{6! 3!}{2^{3} 9!} + \cdot \cdot \cdot \frac{(2 n)! n!}{2^{n} (3 n)!} + \cdot \cdot \cdot$ $S_{n} = \underset{n = 0}{\sum^{\infty}} \frac{Γ (2 n + 1) Γ (n + 1)}{2^{n} Γ (3 n + 1)} = \underset{n = 0}{\sum^{\infty}} \frac{n Γ (2 n + 1) Γ (n)}{2^{n} Γ (3 n + 1)}$ $= \underset{n = 0}{\sum^{\infty}} \frac{n}{2^{n}} β (2 n + 1, n) = \underset{n = 0}{\sum^{\infty}} \frac{n}{2^{n}} \int_{0}^{1} x^{2 n} {(1 - x)}^{n - 1} dx$ $= \int_{0}^{1} {\underset{n = 0}{\sum^{\infty}} [\frac{n}{2^{n}} x^{2 n} {(1 - x)}^{n - 1}]} dx = \int_{0}^{1} {\frac{1}{1 - x} \cdot \underset{n = 0}{\sum^{\infty}} n {(\frac{x^{2} (1 - x)}{2})}^{n}} dx$ $S (t) = \underset{k = 0}{\sum^{\infty}} t^{k} = \frac{1}{1 - t} \Rightarrow S^{'} (t) = \underset{k = 0}{\sum^{\infty}} {kt}^{k - 1} = \frac{1}{{(1 - t)}^{2}}$ ${tS}^{'} (t) = \underset{k = 0}{\sum^{\infty}} {kt}^{k} = \frac{t}{{(1 - t)}^{2}}$ $\underset{n = 0}{\sum^{\infty}} n {(\frac{x^{2} (1 - x)}{2})}^{n} = \frac{\frac{x^{2} (1 - x)}{2}}{{(1 - \frac{x^{2} (1 - x)}{2})}^{2}} = \frac{x^{2} (1 - x)}{2} \cdot \frac{4}{{(2 - x^{2} + x^{3})}^{2}}$ $= \frac{{2 x}^{2} (1 - x)}{{(x^{3} - x^{2} + 2)}^{2}}$ $S_{n} = \int_{0}^{1} {\frac{1}{1 - x} \cdot \frac{{2 x}^{2} (1 - x)}{{(x^{3} - x^{2} + 2)}^{2}}} dx = \int_{0}^{1} \frac{{2 x}^{2} dx}{{(x^{3} - x^{2} + 2)}^{2}}$ $= \int_{0}^{1} \frac{{2 x}^{2}}{{((x + 1) (x^{2} - 2 x + 2))}^{2}}$ $f (x) = 2 {(\frac{x}{(x + 1) (x^{2} - 2 x + 2)})}^{2} = \frac{2}{25} {(\frac{x + 2}{x^{2} - 2 x + 2} - \frac{1}{x + 1})}^{2}$ $= \frac{2}{25} (\frac{x^{2} + 4 x + 4}{{(x^{2} - 2 x + 2)}^{2}} - \frac{2 (x + 2)}{(x + 1) (x^{2} - 2 x + 2)} + \frac{1}{{(x + 1)}^{2}})$ $= \frac{2}{25} (\frac{1}{x^{2} - 2 x + 2} + \frac{6 x + 2}{{(x^{2} - 2 x + 2)}^{2}} - \frac{2}{5 (x + 1)} + \frac{2 (x - 1)}{x^{2} - 2 x + 2} + \frac{1}{{(x + 1)}^{2}})$ $6 x + 2 = 3 (2 x - 2) + 8$ $\int_{0}^{1} f (x) dx = \frac{2}{25} [\tan^{- 1} (x - 1) - \frac{3}{x^{2} - 2 x + 2} + 8 \int \frac{dx}{{({(x - 1)}^{2} + 1)}^{2}}$ ${[- \frac{2 \ln (x + 1)}{5} - \ln (x^{2} - 2 x + 2) - \frac{1}{x + 1}]}_{0}^{1}$ $S_{n} = \frac{2}{25} (- 3 - \frac{2 \ln 2}{5} - \frac{1}{2}) - \frac{2}{25} (- \frac{π}{4} - \frac{3}{2} - \ln 2 - 1) - \frac{2 π}{25}$ $= - \frac{2}{25} + \frac{6 \ln 2}{125} - \frac{6 π}{100}$ $T y p o s$
	Commented by Dwaipayan Shikari last updated on 01/Jan/21

	$\underset{n = 0}{\sum^{\infty}} \frac{1}{2^{n} (\begin{matrix} 2 n \\ n \end{matrix})} = \underset{n = 0}{\sum^{\infty}} \frac{Γ^{2} (n + 1)}{2^{n} Γ (2 n + 1)} = \int_{0}^{1} \underset{n = 0}{\sum^{\infty}} \frac{n}{2^{n}} x^{n} {(1 - x)}^{n - 1} d x$ $= 2 \int_{0}^{1} \frac{x}{{(2 - x (1 - x))}^{2}} d x = \int_{0}^{1} \frac{2 x - 1}{{(x^{2} - x + 2)}^{2}} + \int_{0}^{1} \frac{1}{{(x^{2} - x + 2)}^{2}} d x$ $= \int_{0}^{1} \frac{1}{{((x - \frac{1 + i \sqrt{7}}{2}) (x - \frac{1 - i \sqrt{7}}{2}))}^{2}} d x = - \frac{1}{7} \int_{0}^{1} \frac{1}{{(x - \frac{1 + i \sqrt{7}}{2})}^{2}} + \frac{1}{{(x - \frac{1 - i \sqrt{7}}{2})}^{2}} - \frac{2}{(x^{2} - x + 2)}$ $= \frac{1}{7} \int_{0}^{1} \frac{1}{x^{2} - x + 2} d x + \frac{1}{7} {[. \frac{2 x - 1}{x^{2} - x + 2}]}_{0}^{1}$ $= \frac{8}{7 \sqrt{7}} {t a n}^{- 1} \frac{1}{\sqrt{7}} + \frac{1}{7}$
	Commented by mnjuly1970 last updated on 01/Jan/21

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