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Question Number 199135 by hardmath last updated on 28/Oct/23
a + (1/a) = 3 find: a^5 + (1/a^5 ) = ?
a+1a=3find:a5+1a5=?
Answered by som(math1967) last updated on 28/Oct/23
123
123
Answered by a.lgnaoui last updated on 28/Oct/23
3^5 =a^5 +(1/a^5 )+5(a^3 +(1/a^3 ))+10(a+(1/a)) =a^5 +(1/a^5 )+5(a+(1/a))[a^2 +(1/a^2 )+1]= =a^5 +(1/a^5 )+5(a+(1/a))[(a+(1/a))^2 −1] ⇒a^5 +(1/a^5 )=3^5 −5×24=243−120 a^5 +(1/a^5 )=123
35=a5+1a5+5(a3+1a3)+10(a+1a)=a5+1a5+5(a+1a)[a2+1a2+1]==a5+1a5+5(a+1a)[(a+1a)2−1]⇒a5+1a5=35−5×24=243−120a5+1a5=123
Commented by hardmath last updated on 28/Oct/23
thank you professor
thankyouprofessor
Answered by Rasheed.Sindhi last updated on 28/Oct/23
a + (1/a) = 3 ; a^5 + (1/a^5 ) = ? (a+(1/a))^2 =3^2 a^2 +(1/a^2 )=9−2=7 (a+(1/a))^3 =3^3 a^3 +(1/a^3 )=27−3(a+(1/a))=27−3(3)=18 (a^2 +(1/a^2 ))(a^3 +(1/a^3 ))=a^5 +(1/a^5 )+a+(1/a) (7)(18)=a^5 +(1/a^5 )+3 a^5 +(1/a^5 )=7×18−3=123
a+1a=3;a5+1a5=?(a+1a)2=32a2+1a2=9−2=7(a+1a)3=33a3+1a3=27−3(a+1a)=27−3(3)=18(a2+1a2)(a3+1a3)=a5+1a5+a+1a(7)(18)=a5+1a5+3a5+1a5=7×18−3=123
Answered by Rasheed.Sindhi last updated on 28/Oct/23
a + (1/a) = 3 ; a^5 + (1/a^5 ) = ? •(a+(1/a))^2 =3^2 =9 a^2 +(1/a^2 )=9−2=7 •(a^2 +(1/a^2 ))^2 =7^2 =49 a^4 +(1/a^4 )=49−2=47 •(a^2 +(1/a^2 ))(a+(1/a))=(7)(3)=21 a^3 +(1/a^3 )+a+(1/a)=21 a^3 +(1/a^3 )=21−(a+(1/a))=21−3=18 (a^4 +(1/a^4 ))(a+(1/a))=(47)(3) a^5 +(1/a^5 )+a^3 +(1/a^3 )=141 a^5 +(1/a^5 )+18=141 a^5 +(1/a^5 )=141−18=123
a+1a=3;a5+1a5=?∙(a+1a)2=32=9a2+1a2=9−2=7∙(a2+1a2)2=72=49a4+1a4=49−2=47∙(a2+1a2)(a+1a)=(7)(3)=21a3+1a3+a+1a=21a3+1a3=21−(a+1a)=21−3=18(a4+1a4)(a+1a)=(47)(3)a5+1a5+a3+1a3=141a5+1a5+18=141a5+1a5=141−18=123
Commented by hardmath last updated on 28/Oct/23
thank you professor
thankyouprofessor
Answered by ajfour last updated on 28/Oct/23
a^2 +(1/a^2 )=7 p^n +q^n =(p+q){p^(n−1) −qp^(n−2) +.... ....+p(−q)^(n−2) +(−q)^(n−1) } hence a^5 +(1/a^5 )=(a+(1/a))(a^4 −a^2 +1−(1/a^2 )+(1/a^4 )) ⇒ Q=3(a^4 +(1/a^4 )−a^2 −(1/a^2 )+1) =3(47−7+1) = 3×41=123
a2+1a2=7pn+qn=(p+q){pn−1−qpn−2+….….+p(−q)n−2+(−q)n−1}hencea5+1a5=(a+1a)(a4−a2+1−1a2+1a4)⇒Q=3(a4+1a4−a2−1a2+1)=3(47−7+1)=3×41=123
Answered by Rasheed.Sindhi last updated on 28/Oct/23
a + (1/a) = 3; a^5 + (1/a^5 ) = ? a + (1/a) = 3⇒ { (((1/a)=3−a^★ )),((a^2 =3a−1^(★★) )) :} ^★ (1/a)=3−a (1/a^2 )=(3−a)^2 =a^2 −6a+9 =(3a−1)−6a+9=−3a+8 (1/a^3 )=((−3a+8)/a)=−3+8((1/a)) =−3+8(3−a)=−8a+21 (1/a^4 )=−8+((21)/a)=−8+21(3−a) =−21a+55 (1/a^5 )=−21+((55)/a)=−21+55(3−a) =−21+165−55a=−55a+144 ^(★★) a^2 =3a−1 a^3 =3a^2 −a=3(3a−1)−a=8a−3 a^4 =8a^2 −3a=8(3a−1)−3a=21a−8 a^5 =21a^2 −8a=21(3a−1)−8a=55a−21 a^5 + (1/a^5 ) =(55a−21)+(−55a+144) =123
a+1a=3;a5+1a5=?a+1a=3⇒{1a=3−a★a2=3a−1★★★1a=3−a1a2=(3−a)2=a2−6a+9=(3a−1)−6a+9=−3a+81a3=−3a+8a=−3+8(1a)=−3+8(3−a)=−8a+211a4=−8+21a=−8+21(3−a)=−21a+551a5=−21+55a=−21+55(3−a)=−21+165−55a=−55a+144★★a2=3a−1a3=3a2−a=3(3a−1)−a=8a−3a4=8a2−3a=8(3a−1)−3a=21a−8a5=21a2−8a=21(3a−1)−8a=55a−21a5+1a5=(55a−21)+(−55a+144)=123
Answered by Skabetix last updated on 28/Oct/23
a^2 −3a+1=0 △=b^2 −4ac=9−4=5>0 x_1 and x_2 =((3±(√)5)/2) a^5 +(1/a^5 )=((3±(√5))/2)+(1/((3±(√5))/2))=123
a2−3a+1=0△=b2−4ac=9−4=5>0x1andx2=3±52a5+1a5=3±52+13±52=123

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