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Question Number 128845 by Dwaipayan Shikari last updated on 10/Jan/21

1+(1/2).((1.4)/((5.1!)^2 ))+(1/3).((1.4.6.9)/((5^2 .2!)^2 ))+(1/4).((1.4.6.9.11.14)/((5^3 .3!)^2 ))+...=((b^2 (√((b−(√b))/2)))/(aπ)) Find 5a−8b

1+12.1.4(5.1!)2+13.1.4.6.9(52.2!)2+14.1.4.6.9.11.14(53.3!)2+...=b2b−b2aπFind5a−8b

Answered by mindispower last updated on 10/Jan/21

=1+Σ_(n≥1) ((Π_(k=0) ^(n−1) (1+5k)(4+5k))/((n+1).(5^n .n!)^2 )) =1+Σ_(n≥1) ((5^(2n) Π_(k=0) ^(n−1) (k+(1/5)).Π_(k=0) ^(n−1) (k+(4/5)))/(5^(2n) .(n+1)!.n!)) =1+Σ_(n≥1) ((((1/5))_n .((4/5))_n )/((2)_n )).(((1)^n )/(n!)) =_2 F_1 ((1/5),(4/5);2;[1])=((Γ(2−1)Γ(2))/(Γ(2−(1/5))Γ(2−(4/5))))=(1/(Γ((9/5))Γ((6/5)))) =Γ((9/5))=(4/5)Γ((4/5)) Γ((6/5))=(1/5)Γ((1/5)) S=((25)/(4Γ((1/5))Γ((4/5))))=((25)/(4π))sin((π/5))=((25)/(4π)).(1/4).(√(10−2(√5))) =((25)/(8π)).(√((5−(√5))/2)),b=5,a=8 5a−8b=0

=1+∑n⩾1∏n−1k=0(1+5k)(4+5k)(n+1).(5n.n!)2=1+∑n⩾152n∏n−1k=0(k+15).∏n−1k=0(k+45)52n.(n+1)!.n!=1+∑n⩾1(15)n.(45)n(2)n.(1)nn!=2F1(15,45;2;[1])=Γ(2−1)Γ(2)Γ(2−15)Γ(2−45)=1Γ(95)Γ(65)=Γ(95)=45Γ(45)Γ(65)=15Γ(15)S=254Γ(15)Γ(45)=254πsin(π5)=254π.14.10−25=258π.5−52,b=5,a=85a−8b=0

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