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x-y-1-and-x-2-y-2-2-Find-the-value-of-x-11-y-11-

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Question Number 191527 by MATHEMATICSAM last updated on 25/Apr/23
x + y = 1 and x^2 + y^2 = 2. Find the value of x^(11) + y^(11) .
$${x}\:+\:{y}\:=\:\mathrm{1}\:\mathrm{and}\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:=\:\mathrm{2}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value} \\ $$$$\mathrm{of}\:{x}^{\mathrm{11}} \:+\:{y}^{\mathrm{11}} . \\ $$
Answered by Rasheed.Sindhi last updated on 25/Apr/23
(x+y)^2 =1⇒x^2 +y^2 +2xy=1 ⇒2+2xy=1⇒xy=−(1/2) (x+y)^3 =1⇒x^3 +y^3 +3xy(x+y)=1 x^3 +y^3 +3(−(1/2))(1)=1 •x^3 +y^3 =1+(3/2)=(5/2) (x^2 +y^2 )(x^3 +y^3 )=(2)((5/2)) ⇒x^5 +y^5 +x^2 y^2 (x+y)=5 ⇒x^5 +y^5 +(−(1/2))^2 (1)=5 x^5 +y^5 =5−(1/4)=((19)/4) determinant (((x^5 +y^5 =((19)/4)))).......(i) (x^3 +y^3 )^2 =((25)/4)⇒x^6 +y^6 +2x^3 y^3 =((25)/4) •x^6 +y^6 +2(−(1/2))^3 =((25)/4)⇒x^6 +y^6 =((13)/2) determinant (((x^6 +y^6 =((13)/2))))........(ii) (i)×(ii): x^(11) +y^(11) +x^5 y^5 (x+y)=(((19)/4))(((13)/2))=((247)/8) x^(11) +y^(11) +(−(1/2))^5 (1)=(((19)/4))(((13)/2))=((247)/8) x^(11) +y^(11) =((247)/8)+(1/(32))=((989)/(32)) determinant (((x^(11) +y^(11) =((989)/(32)))))
$$\left({x}+{y}\right)^{\mathrm{2}} =\mathrm{1}\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{xy}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{2}+\mathrm{2}{xy}=\mathrm{1}\Rightarrow{xy}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left({x}+{y}\right)^{\mathrm{3}} =\mathrm{1}\Rightarrow{x}^{\mathrm{3}} +{y}^{\mathrm{3}} +\mathrm{3}{xy}\left({x}+{y}\right)=\mathrm{1} \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} +\mathrm{3}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{1}\right)=\mathrm{1} \\ $$$$\bullet{x}^{\mathrm{3}} +{y}^{\mathrm{3}} =\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} \right)=\left(\mathrm{2}\right)\left(\frac{\mathrm{5}}{\mathrm{2}}\right) \\ $$$$\Rightarrow{x}^{\mathrm{5}} +{y}^{\mathrm{5}} +{x}^{\mathrm{2}} {y}^{\mathrm{2}} \left({x}+{y}\right)=\mathrm{5} \\ $$$$\Rightarrow{x}^{\mathrm{5}} +{y}^{\mathrm{5}} +\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \left(\mathrm{1}\right)=\mathrm{5} \\ $$$$\:\:\:\:{x}^{\mathrm{5}} +{y}^{\mathrm{5}} =\mathrm{5}−\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{19}}{\mathrm{4}} \\ $$$$\begin{array}{|c|}{{x}^{\mathrm{5}} +{y}^{\mathrm{5}} =\frac{\mathrm{19}}{\mathrm{4}}}\\\hline\end{array}…….\left({i}\right) \\ $$$$\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} \right)^{\mathrm{2}} =\frac{\mathrm{25}}{\mathrm{4}}\Rightarrow{x}^{\mathrm{6}} +{y}^{\mathrm{6}} +\mathrm{2}{x}^{\mathrm{3}} {y}^{\mathrm{3}} =\frac{\mathrm{25}}{\mathrm{4}} \\ $$$$\bullet{x}^{\mathrm{6}} +{y}^{\mathrm{6}} +\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} =\frac{\mathrm{25}}{\mathrm{4}}\Rightarrow{x}^{\mathrm{6}} +{y}^{\mathrm{6}} =\frac{\mathrm{13}}{\mathrm{2}} \\ $$$$\begin{array}{|c|}{{x}^{\mathrm{6}} +{y}^{\mathrm{6}} =\frac{\mathrm{13}}{\mathrm{2}}}\\\hline\end{array}……..\left({ii}\right) \\ $$$$\left({i}\right)×\left({ii}\right): \\ $$$${x}^{\mathrm{11}} +{y}^{\mathrm{11}} +{x}^{\mathrm{5}} {y}^{\mathrm{5}} \left({x}+{y}\right)=\left(\frac{\mathrm{19}}{\mathrm{4}}\right)\left(\frac{\mathrm{13}}{\mathrm{2}}\right)=\frac{\mathrm{247}}{\mathrm{8}} \\ $$$${x}^{\mathrm{11}} +{y}^{\mathrm{11}} +\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{5}} \left(\mathrm{1}\right)=\left(\frac{\mathrm{19}}{\mathrm{4}}\right)\left(\frac{\mathrm{13}}{\mathrm{2}}\right)=\frac{\mathrm{247}}{\mathrm{8}} \\ $$$${x}^{\mathrm{11}} +{y}^{\mathrm{11}} =\frac{\mathrm{247}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{32}}=\frac{\mathrm{989}}{\mathrm{32}} \\ $$$$\begin{array}{|c|}{{x}^{\mathrm{11}} +{y}^{\mathrm{11}} =\frac{\mathrm{989}}{\mathrm{32}}}\\\hline\end{array} \\ $$
Answered by mr W last updated on 25/Apr/23
Method II p=e_1 =1 p_2 =e_1 p_1 −2e_2 ⇒2=1−2e_2 ⇒e_2 =−(1/2) p_n =e_1 p_(n−1) −e_2 p_(n−1) ⇒p_n −p_(n−1) −(1/2)p_(n−1) =0 r^2 −r−(1/2)=0 r=((1±(√3))/2) ⇒p_n =(((1+(√3))/2))^n +(((1−(√3))/2))^n examples: p_5 =(((1+(√3))/2))^5 +(((1−(√3))/2))^5 =((19)/4) p_6 =(((1+(√3))/2))^6 +(((1−(√3))/2))^6 =((13)/2) p_(11) =(((1+(√3))/2))^(11) +(((1−(√3))/2))^(11) =((989)/(32)) ✓ p_(15) =(((1+(√3))/2))^(15) +(((1−(√3))/2))^(15) =((13775)/(128)) p_(20) =(((1+(√3))/2))^(20) +(((1−(√3))/2))^(20) =((261575)/(512))
$$\boldsymbol{{Method}}\:\boldsymbol{{II}} \\ $$$${p}={e}_{\mathrm{1}} =\mathrm{1} \\ $$$${p}_{\mathrm{2}} ={e}_{\mathrm{1}} {p}_{\mathrm{1}} −\mathrm{2}{e}_{\mathrm{2}} \:\Rightarrow\mathrm{2}=\mathrm{1}−\mathrm{2}{e}_{\mathrm{2}} \:\Rightarrow{e}_{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${p}_{{n}} ={e}_{\mathrm{1}} {p}_{{n}−\mathrm{1}} −{e}_{\mathrm{2}} {p}_{{n}−\mathrm{1}} \\ $$$$\Rightarrow{p}_{{n}} −{p}_{{n}−\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}{p}_{{n}−\mathrm{1}} =\mathrm{0} \\ $$$${r}^{\mathrm{2}} −{r}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$${r}=\frac{\mathrm{1}\pm\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\Rightarrow{p}_{{n}} =\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{{n}} +\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{{n}} \\ $$$${examples}: \\ $$$${p}_{\mathrm{5}} =\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{5}} +\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{5}} =\frac{\mathrm{19}}{\mathrm{4}} \\ $$$${p}_{\mathrm{6}} =\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{6}} +\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{6}} =\frac{\mathrm{13}}{\mathrm{2}} \\ $$$${p}_{\mathrm{11}} =\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{11}} +\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{11}} =\frac{\mathrm{989}}{\mathrm{32}}\:\checkmark \\ $$$${p}_{\mathrm{15}} =\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{15}} +\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{15}} =\frac{\mathrm{13775}}{\mathrm{128}} \\ $$$${p}_{\mathrm{20}} =\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{20}} +\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{20}} =\frac{\mathrm{261575}}{\mathrm{512}} \\ $$

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