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Question-213796




Question Number 213796 by efronzo1 last updated on 17/Nov/24
Answered by A5T last updated on 17/Nov/24
x_0 =k⇒x_1 =((1+k)/(1−k))⇒x_2 =((−1)/k)⇒x_3 =((k−1)/(1+k))⇒x_4 =((2k)/2)=k  ⇒x_(4n) =k=2022
x0=kx1=1+k1kx2=1kx3=k11+kx4=2k2=kx4n=k=2022
Answered by Frix last updated on 17/Nov/24
n≥0:  x_(4n) =x_0   x_(4n+1) =((1+x_0 )/(1−x_0 ))  x_(4n+2) =−(1/x_0 )  x_(4n+3) =((x_0 −1)/(x_0 +1))  It′s easy...
n0:x4n=x0x4n+1=1+x01x0x4n+2=1x0x4n+3=x01x0+1Itseasy
Answered by golsendro last updated on 17/Nov/24
   x_n  = ((1+x_(n−1) )/(1−x_(n−1) )) ; n−1 ≥ 0    x_1  = ((2023)/(−2021)) , x_2 =((1−((2023)/(2021)))/(1+((2023)/(2021)))) = ((−1)/(2022))   x_3  = ((1−(1/(2022)))/(1+(1/(2022)))) = ((2021)/(2023)) , x_4 = ((1+((2021)/(2023)))/(1−((2021)/(2023))))=2022   ⇒ 4 ∣2024 ; x_(2024)  = x_0 =x_4 =x_8 ...=2022
xn=1+xn11xn1;n10x1=20232021,x2=1202320211+20232021=12022x3=1120221+12022=20212023,x4=1+20212023120212023=202242024;x2024=x0=x4=x8=2022
Answered by mr W last updated on 17/Nov/24
let x_n =tan θ_n   tan θ_(n+1) =x_(n+1) =((1+tan θ_n )/(1−tan θ_n ))=tan (θ_n +(π/4))  ⇒θ_(n+1) =θ_n +(π/4)  ⇒θ_n =θ_(n−1) +(π/4)=θ_(n−2) +((2π)/4)=...=θ_0 +((nπ)/4)  ⇒x_n =tan (θ_0 +((nπ)/4))=((tan θ_0 +tan ((nπ)/4))/(1−tan θ_n tan ((nπ)/4)))  ⇒x_n =((x_0 +tan ((nπ)/4))/(1−x_0 tan ((nπ)/4)))=((2022+tan ((nπ)/4))/(1−2022 tan ((nπ)/4)))  x_(2024) =((2022+tan ((2024π)/4))/(1−2022 tan ((2024π)/4)))            =((2022+0)/(1−2022×0))=2022
letxn=tanθntanθn+1=xn+1=1+tanθn1tanθn=tan(θn+π4)θn+1=θn+π4θn=θn1+π4=θn2+2π4==θ0+nπ4xn=tan(θ0+nπ4)=tanθ0+tannπ41tanθntannπ4xn=x0+tannπ41x0tannπ4=2022+tannπ412022tannπ4x2024=2022+tan2024π412022tan2024π4=2022+012022×0=2022

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