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Question Number 217730 by ArshadS last updated on 19/Mar/25
x+(1/x)=3 ; x^5 +(1/x^5 )=?
$${x}+\frac{\mathrm{1}}{{x}}=\mathrm{3}\:;\:{x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }=? \\ $$
Answered by Wuji last updated on 19/Mar/25
let S_n =x^n +(1/x^n ) S_1 =x+(1/x)=3 S_2 =(x+(1/x))^2 −2 ⇒S_2 =3^2 −2=7 S_3 =(x+(1/x))S_2 −S_1 ⇒S_3 =3×7−3=18 S_4 =(x+(1/x))S_3 −S_2 ⇒S_4 =3×18−7=47 S_5 =(x+(1/x))S_4 −S_3 ⇒S_5 =3×47−18=123 S_5 =x^5 +(1/x^5 )=123
$${let}\:{S}_{{n}} ={x}^{{n}} +\frac{\mathrm{1}}{{x}^{{n}} } \\ $$$${S}_{\mathrm{1}} ={x}+\frac{\mathrm{1}}{{x}}=\mathrm{3} \\ $$$${S}_{\mathrm{2}} =\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}\:\:\Rightarrow{S}_{\mathrm{2}} =\mathrm{3}^{\mathrm{2}} −\mathrm{2}=\mathrm{7} \\ $$$${S}_{\mathrm{3}} =\left({x}+\frac{\mathrm{1}}{{x}}\right){S}_{\mathrm{2}} −{S}_{\mathrm{1}} \:\Rightarrow{S}_{\mathrm{3}} =\mathrm{3}×\mathrm{7}−\mathrm{3}=\mathrm{18} \\ $$$${S}_{\mathrm{4}} =\left({x}+\frac{\mathrm{1}}{{x}}\right){S}_{\mathrm{3}} −{S}_{\mathrm{2}} \:\Rightarrow{S}_{\mathrm{4}} =\mathrm{3}×\mathrm{18}−\mathrm{7}=\mathrm{47} \\ $$$${S}_{\mathrm{5}} =\left({x}+\frac{\mathrm{1}}{{x}}\right){S}_{\mathrm{4}} −{S}_{\mathrm{3}} \:\Rightarrow{S}_{\mathrm{5}} =\mathrm{3}×\mathrm{47}−\mathrm{18}=\mathrm{123} \\ $$$${S}_{\mathrm{5}} ={x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }=\mathrm{123} \\ $$
Commented by ArshadS last updated on 22/Mar/25
Thanks sir!
$${Thanks}\:{sir}! \\ $$
Answered by Rasheed.Sindhi last updated on 19/Mar/25
x+(1/x)=3 ; x^5 +(1/x^5 )=? x+(1/x)=3⇒x^2 =3x−1 ⇒x^4 =9x^2 −6x+1 =9(3x−1)−6x+1 =21x−8 ⇒ determinant ((x^5 ))=21x^2 −8x=21(3x−1)−8x =55x−21 (1/x)=3−x (1/x^2 )=(3−x)^2 =x^2 −6x+9 =(3x−1)−6x+9=−3x+8 (1/x^4 )=(−3x+8)^2 =9x^2 −48x+64 =9(3x−1)−48x+64 =27x−9−48x+64 =−21x+55 determinant (((1/x^5 )))=−21+55((1/x))=−21+55(3−x) =−21+165−55x=−55x+144 determinant (((x^5 +(1/x^5 ))))=(55x−21)+(−55x+144) =123
$${x}+\frac{\mathrm{1}}{{x}}=\mathrm{3}\:;\:{x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }=? \\ $$$${x}+\frac{\mathrm{1}}{{x}}=\mathrm{3}\Rightarrow{x}^{\mathrm{2}} =\mathrm{3}{x}−\mathrm{1} \\ $$$$\:\:\Rightarrow{x}^{\mathrm{4}} =\mathrm{9}{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{9}\left(\mathrm{3}{x}−\mathrm{1}\right)−\mathrm{6}{x}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{21}{x}−\mathrm{8} \\ $$$$\Rightarrow\begin{array}{|c|}{{x}^{\mathrm{5}} }\\\hline\end{array}=\mathrm{21}{x}^{\mathrm{2}} −\mathrm{8}{x}=\mathrm{21}\left(\mathrm{3}{x}−\mathrm{1}\right)−\mathrm{8}{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{55}{x}−\mathrm{21} \\ $$$$\: \\ $$$$\frac{\mathrm{1}}{{x}}=\mathrm{3}−{x} \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\left(\mathrm{3}−{x}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{9} \\ $$$$\:\:\:\:\:\:=\left(\mathrm{3}{x}−\mathrm{1}\right)−\mathrm{6}{x}+\mathrm{9}=−\mathrm{3}{x}+\mathrm{8} \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{4}} }=\left(−\mathrm{3}{x}+\mathrm{8}\right)^{\mathrm{2}} =\mathrm{9}{x}^{\mathrm{2}} −\mathrm{48}{x}+\mathrm{64} \\ $$$$\:\:\:\:\:=\mathrm{9}\left(\mathrm{3}{x}−\mathrm{1}\right)−\mathrm{48}{x}+\mathrm{64} \\ $$$$\:\:\:\:=\mathrm{27}{x}−\mathrm{9}−\mathrm{48}{x}+\mathrm{64} \\ $$$$\:\:\:\:\:=−\mathrm{21}{x}+\mathrm{55} \\ $$$$\begin{array}{|c|}{\frac{\mathrm{1}}{{x}^{\mathrm{5}} }}\\\hline\end{array}=−\mathrm{21}+\mathrm{55}\left(\frac{\mathrm{1}}{{x}}\right)=−\mathrm{21}+\mathrm{55}\left(\mathrm{3}−{x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:=−\mathrm{21}+\mathrm{165}−\mathrm{55}{x}=−\mathrm{55}{x}+\mathrm{144} \\ $$$$\: \\ $$$$\begin{array}{|c|}{{x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }}\\\hline\end{array}=\left(\mathrm{55}{x}−\mathrm{21}\right)+\left(−\mathrm{55}{x}+\mathrm{144}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{123} \\ $$
Commented by ArshadS last updated on 22/Mar/25
Thanks sir!
$${Thanks}\:{sir}! \\ $$
Answered by Ghisom last updated on 20/Mar/25
x^2 −3x+1=0 x^2 =3x−1 x^5 =x(x^2 )^2 =x(3x−1)^2 =x(9x^2 −6x+1)= =x(9(3x−1)−6x+1)=x(21x−8)= =21x^2 −8x=21(3x−1)−8x=55x−21 (1/x^5 )=(1/(55x−21))=(1/(55(3−(1/x))−21))=(x/(144x−55)) 55x−21+(x/(144x−55))= =(((55x−21)(144x−55)+x)/(144x−55))= =((7920x^2 −6048x+1155)/(144x−55))= =((7920(3x−1)−6048x+1155)/(144x−55))= =((17712x−6765)/(144x−55))=123
$${x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{1}=\mathrm{0} \\ $$$$ \\ $$$${x}^{\mathrm{2}} =\mathrm{3}{x}−\mathrm{1} \\ $$$${x}^{\mathrm{5}} ={x}\left({x}^{\mathrm{2}} \right)^{\mathrm{2}} ={x}\left(\mathrm{3}{x}−\mathrm{1}\right)^{\mathrm{2}} ={x}\left(\mathrm{9}{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{1}\right)= \\ $$$$={x}\left(\mathrm{9}\left(\mathrm{3}{x}−\mathrm{1}\right)−\mathrm{6}{x}+\mathrm{1}\right)={x}\left(\mathrm{21}{x}−\mathrm{8}\right)= \\ $$$$=\mathrm{21}{x}^{\mathrm{2}} −\mathrm{8}{x}=\mathrm{21}\left(\mathrm{3}{x}−\mathrm{1}\right)−\mathrm{8}{x}=\mathrm{55}{x}−\mathrm{21} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{5}} }=\frac{\mathrm{1}}{\mathrm{55}{x}−\mathrm{21}}=\frac{\mathrm{1}}{\mathrm{55}\left(\mathrm{3}−\frac{\mathrm{1}}{{x}}\right)−\mathrm{21}}=\frac{{x}}{\mathrm{144}{x}−\mathrm{55}} \\ $$$$ \\ $$$$\mathrm{55}{x}−\mathrm{21}+\frac{{x}}{\mathrm{144}{x}−\mathrm{55}}= \\ $$$$=\frac{\left(\mathrm{55}{x}−\mathrm{21}\right)\left(\mathrm{144}{x}−\mathrm{55}\right)+{x}}{\mathrm{144}{x}−\mathrm{55}}= \\ $$$$=\frac{\mathrm{7920}{x}^{\mathrm{2}} −\mathrm{6048}{x}+\mathrm{1155}}{\mathrm{144}{x}−\mathrm{55}}= \\ $$$$=\frac{\mathrm{7920}\left(\mathrm{3}{x}−\mathrm{1}\right)−\mathrm{6048}{x}+\mathrm{1155}}{\mathrm{144}{x}−\mathrm{55}}= \\ $$$$=\frac{\mathrm{17712}{x}−\mathrm{6765}}{\mathrm{144}{x}−\mathrm{55}}=\mathrm{123} \\ $$
Commented by ArshadS last updated on 22/Mar/25
Thanks sir!
$${Thanks}\:{sir}! \\ $$

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