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x-1-x-1-x-100-1-x-100-x-2020-1-x-2020-




Question Number 77591 by naka3546 last updated on 08/Jan/20
x + (1/x)  =  1  x^(100)  + (1/x^(100) )  =  ?  x^(2020)  + (1/x^(2020) )  =  ?
x+1x=1x100+1x100=?x2020+1x2020=?
Commented by jagoll last updated on 08/Jan/20
(1/x^2 )+x^2 =−1  (1/x^4 )+x^4 =−1  (1/x^8 )+x^8  = −1  (1/x^(2n) )+x^(2n) =−1 so (1/x^(100) )+x^(100) =−1
1x2+x2=11x4+x4=11x8+x8=11x2n+x2n=1so1x100+x100=1
Commented by abdomathmax last updated on 08/Jan/20
x+(1/x)=1 ⇒x^2 +1=x ⇒x^2 −x+1=0  Δ=1−4=−3 ⇒x_1 =((1+i(√3))/2) =e^((iπ)/3)   x_2 =((1−i(√3))/2) =e^(−((iπ)/3))    x=x_1  ⇒x^(100)  +x^(−100) = e^((i100π)/3)  +e^(−((i100π)/3))   =2cos(((100π)/3)) =2cos(((99π+π)/3))=2cos(33π +(π/3))  = 2cos(π+(π/3))=−2×(1/2) =−1  x=x_2  ⇒x^(100)  +x^(−100) = e^(−((i100π)/3)) +e^((i100π)/3) =−1  for x^(2020)  +(1/x^(2020) )  x=x_1  ⇒x^(2020)  +x^(−2020) =e^(i((2020π)/3))  +e^(−((i2020π)/3))   =2cos(((2020π)/3))=2cos(((2019π+π)/3))  =2cos(((2019π)/3)+(π/3))=...
x+1x=1x2+1=xx2x+1=0Δ=14=3x1=1+i32=eiπ3x2=1i32=eiπ3x=x1x100+x100=ei100π3+ei100π3=2cos(100π3)=2cos(99π+π3)=2cos(33π+π3)=2cos(π+π3)=2×12=1x=x2x100+x100=ei100π3+ei100π3=1forx2020+1x2020x=x1x2020+x2020=ei2020π3+ei2020π3=2cos(2020π3)=2cos(2019π+π3)=2cos(2019π3+π3)=
Commented by naka3546 last updated on 08/Jan/20
n= 6  ⇒  the  answer  is  not  −1
n=6theanswerisnot1
Commented by jagoll last updated on 08/Jan/20
how you get 6 sir?
howyouget6sir?
Commented by naka3546 last updated on 08/Jan/20
prove  it  if  the  formula  is  right  for  all  n .
proveitiftheformulaisrightforalln.
Answered by MJS last updated on 08/Jan/20
x=(1/2)±((√3)/2)i ⇒ (1/x)=(1/2)∓((√3)/2)i  z=x^m +(1/x^m )  x^(6n) =1 ⇒ z=2  x^(6n+1) =x ⇒ z=1  x^(6n+2) =−(1/x)=−x^�  ⇒ z=−1  x^(6n+3) =−1 ⇒ z=−2  x^(6n+4) =−x ⇒ z=−1  x^(6n+5) =(1/x)=x^�  ⇒ z=1  100=16×6+4 ⇒ z=−1  2020=336×6+4 ⇒ z=−1
x=12±32i1x=1232iz=xm+1xmx6n=1z=2x6n+1=xz=1x6n+2=1x=x¯z=1x6n+3=1z=2x6n+4=xz=1x6n+5=1x=x¯z=1100=16×6+4z=12020=336×6+4z=1

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