Menu Close

x-1-x-5-x-5-1-x-5-

Tinku Tara 0 Comments
Pin




Question Number 42030 by Akashuac last updated on 17/Aug/18
x+(1/x)=5 x^5 +(1/x^5 )=?
x+1x=5x5+1x5=?
Commented by maxmathsup by imad last updated on 17/Aug/18
x+(1/x)=5 ⇒(x+(1/x))^5 =5^5 ⇒Σ_(k=0) ^5 C_5 ^k x^k ((1/x))^(5−k) =5^5 ⇒ Σ_(k=0) ^5 C_5 ^k x^(2k−5) =5^5 ⇒ x^(−5) + 5 x^(−3) + C_5 ^2 x^(−1) +C_5 ^3 x +5 x^3 +x^5 =5^5 ⇒ x^5 +x^(−5) +5(x^3 +x^(−3) ) +C_5 ^2 (x+x^(−1) ) =5^5 ⇒ x^5 +x^(−5) =5^5 −5(x^3 +x^(−3) ) −C_5 ^2 (x+(1/x)) but (x+(1/x))^3 =5^3 ⇒ x^3 +3x^2 (1/x) +3x (1/x^2 ) +(1/x^3 ) =5^3 ⇒ x^3 +x^(−3) +3(x+(1/x)) =5^3 ⇒x^3 +x^(−3) =5^3 −3×5 we have C_5 ^2 =((5!)/(2!3!)) =((5.4)/2) =10 ⇒x^5 +x^(−5) =5^5 −5(5^3 −15)−10×5 =5^5 −5^4 +5×15 −50 = 5^4 ×4 +25 =4 ×25^2 +25 =4(((100)/4))^2 +25 =((10^4 )/4) +25 =((10000)/4) +25 =2500 +25 =2525 .
x+1x=5⇒(x+1x)5=55⇒∑k=05C5kxk(1x)5−k=55⇒∑k=05C5kx2k−5=55⇒x−5+5x−3+C52x−1+C53x+5x3+x5=55⇒x5+x−5+5(x3+x−3)+C52(x+x−1)=55⇒x5+x−5=55−5(x3+x−3)−C52(x+1x)but(x+1x)3=53⇒x3+3x21x+3x1x2+1x3=53⇒x3+x−3+3(x+1x)=53⇒x3+x−3=53−3×5wehaveC52=5!2!3!=5.42=10⇒x5+x−5=55−5(53−15)−10×5=55−54+5×15−50=54×4+25=4×252+25=4(1004)2+25=1044+25=100004+25=2500+25=2525.
Answered by MrW3 last updated on 17/Aug/18
x^5 +(1/x^5 )= =(x+(1/x))(x^4 −x^3 ×(1/x)+x^2 ×(1/x^2 )−x×(1/x^3 )+(1/x^4 )) =5(x^4 −x^2 +1−(1/x^2 )+(1/x^4 )) =5(x^4 +(1/x^4 )+2x^2 ×(1/x^2 )−x^2 −(1/x^2 )−2x×(1/x)+1−2+2) =5[(x^2 +(1/x^2 ))^2 −(x+(1/x))^2 +1] =5[(x^2 +(1/x^2 )+2x×(1/x)−2)^2 −5^2 +1] =5[{(x+(1/x))^2 −2}^2 −5^2 +1] =5[{5^2 −2}^2 −5^2 +1] =5[23^2 −5^2 +1] =5[28×18+1] =5×505 =2525 in general: =a[(a^2 −2)^2 −a^2 +1] =a[{a(a+1)−2}{(a−1)a−2}+1]
x5+1x5==(x+1x)(x4−x3×1x+x2×1x2−x×1x3+1x4)=5(x4−x2+1−1x2+1x4)=5(x4+1x4+2x2×1x2−x2−1x2−2x×1x+1−2+2)=5[(x2+1x2)2−(x+1x)2+1]=5[(x2+1x2+2x×1x−2)2−52+1]=5[{(x+1x)2−2}2−52+1]=5[{52−2}2−52+1]=5[232−52+1]=5[28×18+1]=5×505=2525ingeneral:=a[(a2−2)2−a2+1]=a[{a(a+1)−2}{(a−1)a−2}+1]
Commented by Akashuac last updated on 17/Aug/18
Thanks you Boss
ThanksyouBoss
Answered by MJS last updated on 17/Aug/18
x+(1/x)=a x^2 −ax+1=0 x=((a±(√(a^2 −4)))/2) (1/x)=(2/(a±(√(a^2 −4))))=((2(a∓(√(a^2 −4))))/(a^2 −(a^2 −4)))=((a∓(√(a^2 −4)))/2) we can write x as p±(√q) and (1/x) as p∓(√q) p∈Q ⇒ x^n +(1/x^n )=[(p+(√q))^n +(p−(√q))^n ]∈Q ∀n∈Z we only need to sum up the p^i ((√q))^j with j=2k n=5 ⇒ x^5 +(1/x^5 )= =2(p^5 +10p^3 q+5pq^2 )= [p=(a/2) q=((a^2 −4)/4)] =a(a^4 −5a^2 +5)= [a=5] =2525
x+1x=ax2−ax+1=0x=a±a2−421x=2a±a2−4=2(a∓a2−4)a2−(a2−4)=a∓a2−42wecanwritexasp±qand1xasp∓qp∈Q⇒xn+1xn=[(p+q)n+(p−q)n]∈Q∀n∈Zweonlyneedtosumupthepi(q)jwithj=2kn=5⇒x5+1x5==2(p5+10p3q+5pq2)=[p=a2q=a2−44]=a(a4−5a2+5)=[a=5]=2525
Commented by Akashuac last updated on 17/Aug/18
thanks boss
thanksboss
Answered by malwaan last updated on 17/Aug/18
by using (x+(1/x))^5 =5^5 (x+(1/x))^3 =5^3 we find x^5 +(1/x^5 )=5^5 −10×5− −5(5^3 −3×5) =2525
byusing(x+1x)5=55(x+1x)3=53wefindx5+1x5=55−10×5−−5(53−3×5)=2525
Commented by Akashuac last updated on 17/Aug/18
thanks boss
thanksboss
Answered by math1967 last updated on 17/Aug/18
(x+(1/x))^2 =25 x^2 +(1/x^2 )=23.........1 x^3 +(1/x^3 )+3(x+(1/x))=125 x^3 +(1/x^3 )=125−15=110..........2 ∴(x^2 +(1/x^2 ))(x^3 +(1/x^3 ))=23×110 x^5 +(1/x^5 )+x+(1/x)=2530 ∴x^5 +(1/x^5 )=2530−5=2525
(x+1x)2=25x2+1x2=23………1x3+1x3+3(x+1x)=125x3+1x3=125−15=110……….2∴(x2+1x2)(x3+1x3)=23×110x5+1x5+x+1x=2530∴x5+1x5=2530−5=2525
Commented by Akashuac last updated on 17/Aug/18
thanks boss
thanksboss

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *