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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=πand∀n⩾0,Un+1=∫0πsin2n+2tdtUn+1=∫0πsint.sin2n+1tdtUn+1=[−cost.sin2n+1t]0π−∫0π(−cost)(2n+1)costsin2ntdtUn+1=(2n+1)∫0πcos2tsin2ntdtUn+1=(2n+1)∫0π(1−sin2t)sin2ntdtUn+1=(2n+1)(Un−Un+1)Un+1=2n+12n+2Un(1)2)(1):Un=2n−12nUn−1=...=∏nk=1(2k−1)2nn!U0Un=∏nk=1(2k−1)2nn!×∏nk=1(2k)2nn!πUn=π(2n)!22nn!2Attention!Audenominateurc′estn!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..c′estmoiquicommetl′erreur..bravoooMrc′estjoliecommesolution.

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