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Question-172448




Question Number 172448 by Shrinava last updated on 27/Jun/22
Answered by Jamshidbek last updated on 28/Jun/22
Let  ((1!!)/(0!))+((3!!)/(1!))+((5!!)/(2!))+...+(((2n+1)!!)/(n!))=a_n   and a_(n+1) −a_n =(((2n+3)!!)/((n+1)!))  Ω=Σ_(n=0) ^∞ a_n ∙((2n+2)/((2n+3)!!))=Σ_(n=0) ^∞ a_n ∙((1/((2n+1)!!))−(1/((2n+3)!!)))=  =Σ_(n=0) ^∞ ((a_n /((2n+1)!!))−(a_n /((2n+3)!!)))=Σ_(n=0) ^∞ (a_n /((2n+1)!!))−Σ_(n=0) ^∞ (a_n /((2n+3)!!))  (1)  Σ_(n=0) ^∞ (a_n /((2n+1)!!))=a_0 +Σ_(n=1) ^∞ (a_n /((2n+1)!!))=a_0 +Σ_(n=0) ^∞ (a_(n+1) /((2n+3)!!))  (2)  From (1) and (2)  Ω=Σ_(n=0) ^∞ (a_n /((2n+1)!!))−Σ_(n=0) ^∞ (a_n /((2n+3)!!))=a_0 +Σ_(n=0) ^∞ (a_(n+1) /((2n+3)!!))−Σ_(n=0) ^∞ (a_n /((2n+3)!!))=  =a_0 +Σ_(n=0) ^∞ ((a_(n+1) −a_n )/((2n+3)!!))=a_0 +Σ_(n=0) ^∞ (((2n+3)!!)/((n+1)!∙(2n+3)!!))=a_0 +Σ_(n=0) ^∞ (1/((n+1)!))=  =a_0 +Σ_(n=1) ^∞ (1/(n!)) ⇒ a_0 =1 ⇒ Ω=1+Σ_(n=1) ^∞ (1/(n!))  e^x =Σ_(n=0) ^∞ (x^n /(n!))=1+Σ_(n=1) ^∞ (x^n /(n!))  Put x=1  e=1+Σ_(n=1) ^∞ (1/(n!)) ⇒ Ω=e  Answer:e  Telegram: @math_undergraduate      =
Let1!!0!+3!!1!+5!!2!++(2n+1)!!n!=anandan+1an=(2n+3)!!(n+1)!Ω=n=0an2n+2(2n+3)!!=n=0an(1(2n+1)!!1(2n+3)!!)==n=0(an(2n+1)!!an(2n+3)!!)=n=0an(2n+1)!!n=0an(2n+3)!!(1)n=0an(2n+1)!!=a0+n=1an(2n+1)!!=a0+n=0an+1(2n+3)!!(2)From(1)and(2)Ω=n=0an(2n+1)!!n=0an(2n+3)!!=a0+n=0an+1(2n+3)!!n=0an(2n+3)!!==a0+n=0an+1an(2n+3)!!=a0+n=0(2n+3)!!(n+1)!(2n+3)!!=a0+n=01(n+1)!==a0+n=11n!a0=1Ω=1+n=11n!ex=n=0xnn!=1+n=1xnn!Putx=1e=1+n=11n!Ω=eAnswer:eTelegram:@math_undergraduate=

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