Question-194331

Question Number 194331 by 073 last updated on 03/Jul/23

Answered by Frix last updated on 03/Jul/23

cos x =C A+B=1 ⇔ B=1−A S_n =C^4 A^n +(C^2 −1)^2 (1−A)^n = =(A^n +(1−A)^n )C^4 −2(1−A)^n C^2 +(1−A)^n It′s easy to show ((S_6 −S_5 )/(4S_4 ))=((S_(10) −S_9 )/(4S_8 ))=((A(A−1))/4) ⇒ Answer is ((A^2 −A)/2) We know A^2 −A−4=0 ⇔ A^2 −A=4 ⇒ Answer is 2

$$\mathrm{cos}\:{x}\:={C} \\ $$$${A}+{B}=\mathrm{1}\:\Leftrightarrow\:{B}=\mathrm{1}−{A} \\ $$$${S}_{{n}} ={C}^{\mathrm{4}} {A}^{{n}} +\left({C}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{1}−{A}\right)^{{n}} = \\ $$$$=\left({A}^{{n}} +\left(\mathrm{1}−{A}\right)^{{n}} \right){C}^{\mathrm{4}} −\mathrm{2}\left(\mathrm{1}−{A}\right)^{{n}} {C}^{\mathrm{2}} +\left(\mathrm{1}−{A}\right)^{{n}} \\ $$$$\mathrm{It}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{show} \\ $$$$\frac{{S}_{\mathrm{6}} −{S}_{\mathrm{5}} }{\mathrm{4}{S}_{\mathrm{4}} }=\frac{{S}_{\mathrm{10}} −{S}_{\mathrm{9}} }{\mathrm{4}{S}_{\mathrm{8}} }=\frac{{A}\left({A}−\mathrm{1}\right)}{\mathrm{4}}\:\Rightarrow \\ $$$$\mathrm{Answer}\:\mathrm{is}\:\frac{{A}^{\mathrm{2}} −{A}}{\mathrm{2}} \\ $$$$\mathrm{We}\:\mathrm{know}\:{A}^{\mathrm{2}} −{A}−\mathrm{4}=\mathrm{0}\:\Leftrightarrow\:{A}^{\mathrm{2}} −{A}=\mathrm{4}\:\Rightarrow \\ $$$$\mathrm{Answer}\:\mathrm{is}\:\mathrm{2} \\ $$

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Question-194331

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