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Question Number 219492 by hardmath last updated on 26/Apr/25
a + b = 1 a^2 + b^2 = 2 Find: a^(11) + b^(11) = ?
$$\mathrm{a}\:+\:\mathrm{b}\:=\:\mathrm{1} \\ $$$$\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} \:=\:\mathrm{2} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{a}^{\mathrm{11}} \:+\:\mathrm{b}^{\mathrm{11}} \:=\:? \\ $$
Answered by mr W last updated on 27/Apr/25
p_1 =e_1 =1 p_2 =1×1−2e_2 =2 ⇒e_2 =−(1/2) p_n −e_1 p_(n−1) +e_2 p_(n−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 p_(11) =((989)/(32))
$${p}_{\mathrm{1}} ={e}_{\mathrm{1}} =\mathrm{1} \\ $$$${p}_{\mathrm{2}} =\mathrm{1}×\mathrm{1}−\mathrm{2}{e}_{\mathrm{2}} =\mathrm{2}\:\Rightarrow{e}_{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${p}_{{n}} −{e}_{\mathrm{1}} {p}_{{n}−\mathrm{1}} +{e}_{\mathrm{2}} {p}_{{n}−\mathrm{2}} =\mathrm{0} \\ $$$${r}^{\mathrm{2}} −{r}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow{r}=\frac{\mathrm{1}\pm\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${p}_{{n}} ={a}^{{n}} +{b}^{{n}} =\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{{n}} +\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{{n}} \\ $$$${p}_{\mathrm{11}} =\frac{\mathrm{989}}{\mathrm{32}} \\ $$
Answered by mr W last updated on 27/Apr/25
(a+b)^2 =a^2 +b^2 +2ab ⇒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^2 +b^2 )^2 =a^4 +b^4 +2(ab)^2 ⇒a^4 +b^4 =2^2 −2×(−(1/2))^2 =(7/2) (a^4 +b^4 )(a+b)=a^5 +b^5 +ab(a^3 +b^3 ) ⇒a^5 +b^5 =(7/2)×1−(−(1/2))×(5/2)=((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) +(a+b)(ab)^5 ⇒a^(11) +b^(11) =((19)/4)×((13)/2)−1×(−(1/2))^5 =((989)/(32))
$$\left({a}+{b}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab}\: \\ $$$$\Rightarrow{ab}=\frac{\mathrm{1}^{\mathrm{2}} −\mathrm{2}}{\mathrm{2}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left({a}+{b}\right)^{\mathrm{3}} ={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +\mathrm{3}{ab}\left({a}+{b}\right) \\ $$$$\Rightarrow{a}^{\mathrm{3}} +{b}^{\mathrm{3}} =\mathrm{1}^{\mathrm{3}} −\mathrm{3}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)×\mathrm{1}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} ={a}^{\mathrm{4}} +{b}^{\mathrm{4}} +\mathrm{2}\left({ab}\right)^{\mathrm{2}} \: \\ $$$$\Rightarrow{a}^{\mathrm{4}} +{b}^{\mathrm{4}} =\mathrm{2}^{\mathrm{2}} −\mathrm{2}×\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\left({a}^{\mathrm{4}} +{b}^{\mathrm{4}} \right)\left({a}+{b}\right)={a}^{\mathrm{5}} +{b}^{\mathrm{5}} +{ab}\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} \right) \\ $$$$\Rightarrow{a}^{\mathrm{5}} +{b}^{\mathrm{5}} =\frac{\mathrm{7}}{\mathrm{2}}×\mathrm{1}−\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)×\frac{\mathrm{5}}{\mathrm{2}}=\frac{\mathrm{19}}{\mathrm{4}} \\ $$$$\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} \right)^{\mathrm{2}} ={a}^{\mathrm{6}} +{b}^{\mathrm{6}} +\mathrm{2}\left({ab}\right)^{\mathrm{3}} \\ $$$$\Rightarrow{a}^{\mathrm{6}} +{b}^{\mathrm{6}} =\left(\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{2}×\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} =\frac{\mathrm{13}}{\mathrm{2}} \\ $$$$\left({a}^{\mathrm{5}} +{b}^{\mathrm{5}} \right)\left({a}^{\mathrm{6}} +{b}^{\mathrm{6}} \right)={a}^{\mathrm{11}} +{b}^{\mathrm{11}} +\left({a}+{b}\right)\left({ab}\right)^{\mathrm{5}} \\ $$$$\Rightarrow{a}^{\mathrm{11}} +{b}^{\mathrm{11}} =\frac{\mathrm{19}}{\mathrm{4}}×\frac{\mathrm{13}}{\mathrm{2}}−\mathrm{1}×\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{5}} =\frac{\mathrm{989}}{\mathrm{32}} \\ $$
Commented by hardmath last updated on 27/Apr/25
thank you very much my dear professor
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{professor} \\ $$
Answered by Rasheed.Sindhi last updated on 03/May/25
a + b = 1...(i) a^2 + b^2 = 2...(ii) a^(11) + b^(11) = ? (i)⇒a=1−b (ii)⇒(1−b)^2 +b^2 =2 2b^2 −2b−1=0 b^2 =b+0.5 b^4 =b^2 +b+0.25 =b+0.5+b+0.25 =2b+0.75 b^8 =4b^2 +3b+0.5625 =4(b+0.5)+3b+0.5625 =7b+2.5625 b^(10) =b^8 .b^2 =(7b+2.5625)(b+0.5) =7b^2 +6.0625b+1.28125 =7(b+0.5)+6.0625b+1.28125 =13.0625b+4.78125 b^(11) =13.0625b^2 +4.78125b =13.0625(b+0.5)+4.78125b =17.84375b+6.53125 Similarly, a^(11) =17.84375a+6.53125 a^(11) +b^(11) =17.84375(a+b)+13.0625 =17.84375(1)+13.0625 =30.90625=((989)/(32))
$$\mathrm{a}\:+\:\mathrm{b}\:=\:\mathrm{1}…\left(\mathrm{i}\right) \\ $$$$\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} \:=\:\mathrm{2}…\left(\mathrm{ii}\right) \\ $$$$\mathrm{a}^{\mathrm{11}} \:+\:\mathrm{b}^{\mathrm{11}} \:=\:? \\ $$$$\left(\mathrm{i}\right)\Rightarrow\mathrm{a}=\mathrm{1}−\mathrm{b} \\ $$$$\:\left(\mathrm{ii}\right)\Rightarrow\left(\mathrm{1}−\mathrm{b}\right)^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} =\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{2b}^{\mathrm{2}} −\mathrm{2b}−\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{b}^{\mathrm{2}} =\mathrm{b}+\mathrm{0}.\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{b}^{\mathrm{4}} =\mathrm{b}^{\mathrm{2}} +\mathrm{b}+\mathrm{0}.\mathrm{25} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{b}+\mathrm{0}.\mathrm{5}+\mathrm{b}+\mathrm{0}.\mathrm{25} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2b}+\mathrm{0}.\mathrm{75} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{b}^{\mathrm{8}} =\mathrm{4b}^{\mathrm{2}} +\mathrm{3b}+\mathrm{0}.\mathrm{5625} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{4}\left(\mathrm{b}+\mathrm{0}.\mathrm{5}\right)+\mathrm{3b}+\mathrm{0}.\mathrm{5625} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{7b}+\mathrm{2}.\mathrm{5625} \\ $$$$\:\:\mathrm{b}^{\mathrm{10}} =\mathrm{b}^{\mathrm{8}} .\mathrm{b}^{\mathrm{2}} =\left(\mathrm{7b}+\mathrm{2}.\mathrm{5625}\right)\left(\mathrm{b}+\mathrm{0}.\mathrm{5}\right) \\ $$$$\:\:\:\:\:\:\:=\mathrm{7b}^{\mathrm{2}} +\mathrm{6}.\mathrm{0625b}+\mathrm{1}.\mathrm{28125} \\ $$$$\:\:\:\:\:\:=\mathrm{7}\left(\mathrm{b}+\mathrm{0}.\mathrm{5}\right)+\mathrm{6}.\mathrm{0625b}+\mathrm{1}.\mathrm{28125} \\ $$$$\:\:\:\:\:\:=\mathrm{13}.\mathrm{0625b}+\mathrm{4}.\mathrm{78125} \\ $$$$\:\mathrm{b}^{\mathrm{11}} =\mathrm{13}.\mathrm{0625b}^{\mathrm{2}} +\mathrm{4}.\mathrm{78125b} \\ $$$$\:\:\:\:=\mathrm{13}.\mathrm{0625}\left(\mathrm{b}+\mathrm{0}.\mathrm{5}\right)+\mathrm{4}.\mathrm{78125b} \\ $$$$\:\:\:=\mathrm{17}.\mathrm{84375b}+\mathrm{6}.\mathrm{53125} \\ $$$$\: \\ $$$$\:\mathrm{Similarly}, \\ $$$$\:\:\:\mathrm{a}^{\mathrm{11}} =\mathrm{17}.\mathrm{84375a}+\mathrm{6}.\mathrm{53125} \\ $$$$\: \\ $$$$\mathrm{a}^{\mathrm{11}} +\mathrm{b}^{\mathrm{11}} =\mathrm{17}.\mathrm{84375}\left(\mathrm{a}+\mathrm{b}\right)+\mathrm{13}.\mathrm{0625} \\ $$$$\:\:\:\:\:\:\:=\mathrm{17}.\mathrm{84375}\left(\mathrm{1}\right)+\mathrm{13}.\mathrm{0625} \\ $$$$\:\:\:\:\:\:=\mathrm{30}.\mathrm{90625}=\frac{\mathrm{989}}{\mathrm{32}} \\ $$

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