nice-mathematics-prove-that-n-1-2n-n-2-2n-1-2-4n-1-2-pi-m-n-july-1970- Tinku Tara June 4, 2023 Differentiation 0 Comments FacebookTweetPin Question Number 114735 by mnjuly1970 last updated on 20/Sep/20 ….nicemathematics…provethat::∑∞n=1[(2nn)]2(2n−1)24n=1−2πYou can't use 'macro parameter character #' in math modeYou can't use 'macro parameter character #' in math mode Answered by maths mind last updated on 20/Sep/20 ∑2n⩾1[(2nn)]2(2n−1)24n=Σ(2n!n!.n!)2(2n−1)24n=S=1+∑n⩾2(2n!n!.n!)2(2n−1)24n∣2n!=2n.∏n−1k=0(2k+1)=22nn!.∏n−1k=0(k+12)∑∞n⩾1[(2nn)]2(2n−1)24n=∑n⩾1(2n!n!.n!)2(2n−1)24n=∑n⩾124nn!2∏n−1k=0(k+12)2(n!)4(2n−1)24n∑n⩾1∏n−1k=0(k+12)2(2n−1)(n!)2.1=14+∑n⩾2∏n−1k=0(k+12)2(2n−1)n!2.1=14+∑n⩾2∏n−1k=0(k+12)∏n−2k=0(k+12)(n−1+12)(2n−1)n!.(1)nn!=14+∑n⩾2∏n−1k=0(12+k)∏n−2k=0(k+12)(2n−1)2(2n−1)n!.1nn!∏n−2k=0(k+12)=−2∏n−1k=0(k−12)=−2(−12)nn!=∏n−1k=0(1+k)=(1)nsoWegetS=14+∑n⩾2(12)n.−2(−12)n2(1)n.1nn!recallthat2F1(a,b;c;x)=1+∑n⩾1anbncnxnn!⇒S=14−∑n⩾2(12)n(−12)n(1)n=14−(2F1(12,−12;1;1)−1−(12)1(−12)1(1)1.11!)=14−(2F1(12,−12;1;1)−1+14)=1−2F1(12;−12;1;1)weusegausseHypergeometriquetheorem2F1(a,b;c,1)=Γ(c)Γ(c−b−a)Γ(c−a)Γ(c−b)thisisjustresultofβ(b,c−b)2F1(a,b;c;z)=∫01xb−1(1−x)c−b+1(1−zx)−adxsoweget2F1(12,−12;1;1)=Γ(1)Γ(1+12−12)Γ(12)Γ(32)=Γ(1)Γ(1)12Γ(12)2=2πS=1−2F1(12,−12;1;1)=1−2π Commented by mnjuly1970 last updated on 21/Sep/20 verynicethankyousir.. Commented by maths mind last updated on 21/Sep/20 withepleasursir Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: 3-4x-5-3-5-1-4-3-x-5-pls-what-is-the-explanation-behind-the-fact-that-i-always-have-to-factorize-to-get-the-form-1-b-s-any-time-i-am-solving-for-binomial-with-negatuve-powNext Next post: Question-180268 Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.