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Question Number 210591 by Tawa11 last updated on 13/Aug/24
If a + b = 1 a^2 + b^2 = 2 then, a^(11) + b^(11) = ??
$$\mathrm{If} \\ $$$$\:\:\:\:\mathrm{a}\:\:+\:\:\mathrm{b}\:\:=\:\:\mathrm{1} \\ $$$$\:\:\:\mathrm{a}^{\mathrm{2}} \:\:+\:\:\mathrm{b}^{\mathrm{2}} \:\:=\:\:\mathrm{2} \\ $$$$\mathrm{then},\:\:\:\:\:\mathrm{a}^{\mathrm{11}} \:\:+\:\:\mathrm{b}^{\mathrm{11}} \:\:=\:\:?? \\ $$
Answered by mr W last updated on 13/Aug/24
p_1 =e_1 =1 p_1 ^2 =2+2e_2 =1 ⇒e_2 =−(1/2) p_n =e_1 p_(n−1) −e_2 p_(n−2) p_n −p_(n−1) −(p_(n−2) /2)=0 r^2 −r−(1/2)=0 ⇒r=((1±(√3))/2) ⇒p_n =a^n +b^n =(((1+(√3))/2))^n +(((1−(√3))/2))^n a^(11) +b^(11) =(((1+(√3))/2))^(11) +(((1−(√3))/2))^(11) =((989)/(32))
$${p}_{\mathrm{1}} ={e}_{\mathrm{1}} =\mathrm{1} \\ $$$${p}_{\mathrm{1}} ^{\mathrm{2}} =\mathrm{2}+\mathrm{2}{e}_{\mathrm{2}} =\mathrm{1}\:\Rightarrow{e}_{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${p}_{{n}} ={e}_{\mathrm{1}} {p}_{{n}−\mathrm{1}} −{e}_{\mathrm{2}} {p}_{{n}−\mathrm{2}} \\ $$$${p}_{{n}} −{p}_{{n}−\mathrm{1}} −\frac{{p}_{{n}−\mathrm{2}} }{\mathrm{2}}=\mathrm{0} \\ $$$${r}^{\mathrm{2}} −{r}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow{r}=\frac{\mathrm{1}\pm\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\Rightarrow{p}_{{n}} ={a}^{{n}} +{b}^{{n}} =\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{{n}} +\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{{n}} \\ $$$${a}^{\mathrm{11}} +{b}^{\mathrm{11}} =\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{11}} +\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{11}} =\frac{\mathrm{989}}{\mathrm{32}} \\ $$
Commented by Tawa11 last updated on 13/Aug/24
Wow. Thanks sir. I appreciate.
$$\mathrm{Wow}.\:\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}. \\ $$
Answered by mr W last updated on 13/Aug/24
a^2 +b^2 +2ab=(a+b)^2 ⇒ab=((1^2 −2)/2)=−(1/2) (a+b)^3 =a^3 +b^3 +3ab(a+b) ⇒a^3 +b^3 =1^3 −3(−(1/2))×1=(5/2) (a^3 +b^3 )(a^2 +b^2 )=a^5 +b^5 +(ab)^2 (a+b) ⇒a^5 +b^5 =(5/2)×2−(−(1/2))^2 ×1=((19)/4) (a^3 +b^3 )^2 =a^6 +b^6 +2(ab)^3 ⇒a^6 +b^6 =((5/2))^2 −2(−(1/2))^3 =((13)/2) (a^5 +b^5 )(a^6 +b^6 )=a^(11) +b^(11) +(ab)^5 (a+b) ⇒a^(11) +b^(11) =((19)/4)×((13)/2)−(−(1/2))^5 ×1=((989)/(32)) ✓
$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab}=\left({a}+{b}\right)^{\mathrm{2}} \:\Rightarrow{ab}=\frac{\mathrm{1}^{\mathrm{2}} −\mathrm{2}}{\mathrm{2}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left({a}+{b}\right)^{\mathrm{3}} ={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +\mathrm{3}{ab}\left({a}+{b}\right)\: \\ $$$$\Rightarrow{a}^{\mathrm{3}} +{b}^{\mathrm{3}} =\mathrm{1}^{\mathrm{3}} −\mathrm{3}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)×\mathrm{1}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} \right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)={a}^{\mathrm{5}} +{b}^{\mathrm{5}} +\left({ab}\right)^{\mathrm{2}} \left({a}+{b}\right) \\ $$$$\Rightarrow{a}^{\mathrm{5}} +{b}^{\mathrm{5}} =\frac{\mathrm{5}}{\mathrm{2}}×\mathrm{2}−\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} ×\mathrm{1}=\frac{\mathrm{19}}{\mathrm{4}} \\ $$$$\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} \right)^{\mathrm{2}} ={a}^{\mathrm{6}} +{b}^{\mathrm{6}} +\mathrm{2}\left({ab}\right)^{\mathrm{3}} \\ $$$$\Rightarrow{a}^{\mathrm{6}} +{b}^{\mathrm{6}} =\left(\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} =\frac{\mathrm{13}}{\mathrm{2}} \\ $$$$\left({a}^{\mathrm{5}} +{b}^{\mathrm{5}} \right)\left({a}^{\mathrm{6}} +{b}^{\mathrm{6}} \right)={a}^{\mathrm{11}} +{b}^{\mathrm{11}} +\left({ab}\right)^{\mathrm{5}} \left({a}+{b}\right) \\ $$$$\Rightarrow{a}^{\mathrm{11}} +{b}^{\mathrm{11}} =\frac{\mathrm{19}}{\mathrm{4}}×\frac{\mathrm{13}}{\mathrm{2}}−\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{5}} ×\mathrm{1}=\frac{\mathrm{989}}{\mathrm{32}}\:\checkmark \\ $$
Commented by Tawa11 last updated on 13/Aug/24
Thanks sir. I appreciate.
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}. \\ $$
Answered by ajfour last updated on 14/Aug/24
a(1+t)=1 a^2 (1+t^2 )=2 a^(11) (1+t^(11) )=x t=(b/a)=2ab+2=1 ⇒ ab=−(1/2) a^2 t=−(1/2) a^2 (1+t)^2 =1 (((1+t)^2 )/t)=−2 ⇒ t^2 +4t+4=3 t=−2±(√3) a=(1/(1+t)) a^(11) +b^(11) =((1+(−2±(√3))^(11) )/((−1±(√3))^(11) ))
$${a}\left(\mathrm{1}+{t}\right)=\mathrm{1} \\ $$$${a}^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)=\mathrm{2} \\ $$$${a}^{\mathrm{11}} \left(\mathrm{1}+{t}^{\mathrm{11}} \right)={x} \\ $$$${t}=\frac{{b}}{{a}}=\mathrm{2}{ab}+\mathrm{2}=\mathrm{1}\:\:\:\Rightarrow\:\:{ab}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} {t}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} \left(\mathrm{1}+{t}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\frac{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }{{t}}=−\mathrm{2} \\ $$$$\Rightarrow\:\:\:{t}^{\mathrm{2}} +\mathrm{4}{t}+\mathrm{4}=\mathrm{3} \\ $$$${t}=−\mathrm{2}\pm\sqrt{\mathrm{3}} \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{1}+{t}} \\ $$$${a}^{\mathrm{11}} +{b}^{\mathrm{11}} =\frac{\mathrm{1}+\left(−\mathrm{2}\pm\sqrt{\mathrm{3}}\right)^{\mathrm{11}} }{\left(−\mathrm{1}\pm\sqrt{\mathrm{3}}\right)^{\mathrm{11}} } \\ $$
Commented by Tawa11 last updated on 14/Aug/24
Thanks sir. I appreciate.
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}. \\ $$

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