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Question Number 151859 by puissant last updated on 23/Aug/21

Commented by OlafThorendsen last updated on 23/Aug/21

1)  U_n  = ∫_0 ^π sin^(2n) t dt  U_0  = ∫_0 ^π dt = π and   ∀n≥0, U_(n+1)  = ∫_0 ^π sin^(2n+2) t dt  U_(n+1)  = ∫_0 ^π sint.sin^(2n+1) t dt  U_(n+1)  = [−cost.sin^(2n+1) t]_0 ^π   −∫_0 ^π (−cost)(2n+1)costsin^(2n) t dt  U_(n+1)  = (2n+1)∫_0 ^π cos^2 tsin^(2n) t dt  U_(n+1)  = (2n+1)∫_0 ^π (1−sin^2 t)sin^(2n) t dt  U_(n+1)  = (2n+1)(U_n −U_(n+1) )  U_(n+1)  = ((2n+1)/(2n+2))U_n      (1)    2)  (1) : U_n  = ((2n−1)/(2n))U_(n−1)  = ... = ((Π_(k=1) ^n (2k−1))/(2^n n!))U_0   U_n  = ((Π_(k=1) ^n (2k−1))/(2^n n!))×((Π_(k=1) ^n (2k))/(2^n n!))π  U_n  = ((π(2n)!)/(2^(2n) n!^2 ))  Attention !  Au denominateur c′est n!^2   et non n!

1)Un=0πsin2ntdtU0=0πdt=πandn0,Un+1=0πsin2n+2tdtUn+1=0πsint.sin2n+1tdtUn+1=[cost.sin2n+1t]0π0π(cost)(2n+1)costsin2ntdtUn+1=(2n+1)0πcos2tsin2ntdtUn+1=(2n+1)0π(1sin2t)sin2ntdtUn+1=(2n+1)(UnUn+1)Un+1=2n+12n+2Un(1)2)(1):Un=2n12nUn1=...=nk=1(2k1)2nn!U0Un=nk=1(2k1)2nn!×nk=1(2k)2nn!πUn=π(2n)!22nn!2Attention!Audenominateurcestn!2etnonn!

Commented by puissant last updated on 23/Aug/21

oui Mr vous connaissez..  c′est moi qui commet l′erreur..  bravooo Mr c′est jolie comme solution.

ouiMrvousconnaissez..cestmoiquicommetlerreur..bravoooMrcestjoliecommesolution.

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