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Bobhans-1-Let-n-be-a-positive-integer-and-let-x-and-y-be-positive-real-number-such-that-x-n-y-n-1-Prove-that-k-1-n-1-x-2k-1-x-4k-k-1-n-1-y-2k-1-




Question Number 108818 by bobhans last updated on 19/Aug/20
  ((Bobhans)/Δ)  (1)Let n be a positive integer, and let x and y be positive real   number such that x^n  + y^n  = 1 . Prove that   (Σ_(k = 1) ^n ((1+x^(2k) )/(1+x^(4k) )) )(Σ_(k = 1) ^n ((1+y^(2k) )/(1+y^(4k) )) ) < (1/((1−x)(1−y)))  (2) All the letters of the word ′EAMCOT ′ are arranged in different    possible ways. The number of such arrangement in which   no two vowels are adjacent to each other is ___
\boldsymbolBob\boldsymbolhansΔ(1)Letnbeapositiveinteger,andletxandybepositiverealnumbersuchthatxn+yn=1.Provethat(nk=11+x2k1+x4k)(nk=11+y2k1+y4k)<1(1x)(1y)(2)AllthelettersofthewordEAMCOTarearrangedindifferentpossibleways.Thenumberofsucharrangementinwhichnotwovowelsareadjacenttoeachotheris___
Answered by john santu last updated on 19/Aug/20
    ((⊸JS⊸)/♦)  (2) we note that there are 3 consonant and  3 vowels A,E and O. Since no two vowels   have tobe together , the possible choice for   vowels are the places marked as ′♦′  ♦M♦C♦T♦, these vowels can be  arranged in P_3 ^( 4)  ways, 3 consonant can be   arranged in 3! ways. Hence the required  number of ways is 3! ×P _3^4   = 6×((4!)/((4−3)!)) = 6×24 = 144
JS(2)wenotethatthereare3consonantand3vowelsA,EandO.Sincenotwovowelshavetobetogether,thepossiblechoiceforvowelsaretheplacesmarkedasMCT,thesevowelscanbearrangedinP34ways,3consonantcanbearrangedin3!ways.Hencetherequirednumberofwaysis3!×P34=6×4!(43)!=6×24=144

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