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advanced-mathematics-prove-that-n-0-1-2-n-3n-n-3-125-11pi-6-2log-2-45-




Question Number 127725 by mnjuly1970 last updated on 01/Jan/21
       ... advanced  mathematics...    prove  that ::           Σ_(n=0) ^∞ (1/(2^n  (((3n)),((  n)) ))) =^(???) (3/(125))(((11π)/6)−2log(2)+45)
advancedmathematicsprovethat::n=012n(3nn)=???3125(11π62log(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+2!23!+4!2!226!+6!3!239!+(2n)!n!2n(3n)!+Sn=n=0Γ(2n+1)Γ(n+1)2nΓ(3n+1)=n=0nΓ(2n+1)Γ(n)2nΓ(3n+1)=n=0n2nβ(2n+1,n)=n=0n2n01x2n(1x)n1dx=01{n=0[n2nx2n(1x)n1]}dx=01{11xn=0n(x2(1x)2)n}dxS(t)=k=0tk=11tS(t)=k=0ktk1=1(1t)2tS(t)=k=0ktk=t(1t)2n=0n(x2(1x)2)n=x2(1x)2(1x2(1x)2)2=x2(1x)24(2x2+x3)2=2x2(1x)(x3x2+2)2Sn=01{11x2x2(1x)(x3x2+2)2}dx=012x2dx(x3x2+2)2=012x2((x+1)(x22x+2))2f(x)=2(x(x+1)(x22x+2))2=225(x+2x22x+21x+1)2=225(x2+4x+4(x22x+2)22(x+2)(x+1)(x22x+2)+1(x+1)2)=225(1x22x+2+6x+2(x22x+2)225(x+1)+2(x1)x22x+2+1(x+1)2)6x+2=3(2x2)+801f(x)dx=225[tan1(x1)3x22x+2+8dx((x1)2+1)2[2ln(x+1)5ln(x22x+2)1x+1]01Sn=225(32ln2512)225(π432ln21)2π25=225+6ln21256π100Typos
Commented by Dwaipayan Shikari last updated on 01/Jan/21
Σ_(n=0) ^∞ (1/(2^n  (((2n)),(n) )))=Σ_(n=0) ^∞ ((Γ^2 (n+1))/(2^n Γ(2n+1)))=∫_0 ^1 Σ_(n=0) ^∞ (n/2^n )x^n (1−x)^(n−1) dx  =2∫_0 ^1 (x/((2−x(1−x))^2 ))dx=∫_0 ^1 ((2x−1)/((x^2 −x+2)^2 ))+∫_0 ^1 (1/((x^2 −x+2)^2 ))dx  =∫_0 ^1 (1/(((x−((1+i(√7))/2))(x−((1−i(√7))/2)))^2 ))dx=−(1/7)∫_0 ^1 (1/((x−((1+i(√7))/2))^2 ))+(1/((x−((1−i(√7))/2))^2 ))−(2/((x^2 −x+2)))  =(1/7)∫_0 ^1 (1/(x^2 −x+2))dx+(1/7)[.((2x−1)/(x^2 −x+2))]_0 ^1   =(8/(7(√7)))tan^(−1) (1/( (√7)))+(1/7)
n=012n(2nn)=n=0Γ2(n+1)2nΓ(2n+1)=01n=0n2nxn(1x)n1dx=201x(2x(1x))2dx=012x1(x2x+2)2+011(x2x+2)2dx=011((x1+i72)(x1i72))2dx=17011(x1+i72)2+1(x1i72)22(x2x+2)=17011x2x+2dx+17[.2x1x2x+2]01=877tan117+17
Commented by mnjuly1970 last updated on 01/Jan/21
nice vety nicr
nicevetynicr

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