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 )=?
$$\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}=\mathrm{5}\:\:\:\:\mathrm{x}^{\mathrm{5}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{5}} }=? \\ $$
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 .
$${x}+\frac{\mathrm{1}}{{x}}=\mathrm{5}\:\Rightarrow\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{5}} \:=\mathrm{5}^{\mathrm{5}} \:\Rightarrow\sum_{{k}=\mathrm{0}} ^{\mathrm{5}} \:\:{C}_{\mathrm{5}} ^{{k}} \:\:{x}^{{k}} \left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{5}−{k}} \:=\mathrm{5}^{\mathrm{5}} \:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{0}} ^{\mathrm{5}} \:{C}_{\mathrm{5}} ^{{k}} \:{x}^{\mathrm{2}{k}−\mathrm{5}} \:\:=\mathrm{5}^{\mathrm{5}} \:\Rightarrow\:\:{x}^{−\mathrm{5}} \:+\:\mathrm{5}\:{x}^{−\mathrm{3}} \:\:+\:{C}_{\mathrm{5}} ^{\mathrm{2}} \:{x}^{−\mathrm{1}} \:\:+{C}_{\mathrm{5}} ^{\mathrm{3}} {x}\:\:+\mathrm{5}\:{x}^{\mathrm{3}} \:\:+{x}^{\mathrm{5}} \:=\mathrm{5}^{\mathrm{5}} \:\Rightarrow \\ $$$${x}^{\mathrm{5}} \:+{x}^{−\mathrm{5}} \:\:+\mathrm{5}\left({x}^{\mathrm{3}} +{x}^{−\mathrm{3}} \right)\:\:+{C}_{\mathrm{5}} ^{\mathrm{2}} \left({x}+{x}^{−\mathrm{1}} \right)\:=\mathrm{5}^{\mathrm{5}} \:\Rightarrow \\ $$$${x}^{\mathrm{5}} \:+{x}^{−\mathrm{5}} \:=\mathrm{5}^{\mathrm{5}} \:−\mathrm{5}\left({x}^{\mathrm{3}} +{x}^{−\mathrm{3}} \right)\:−{C}_{\mathrm{5}} ^{\mathrm{2}} \left({x}+\frac{\mathrm{1}}{{x}}\right)\:\:{but} \\ $$$$\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} \:=\mathrm{5}^{\mathrm{3}} \:\Rightarrow\:{x}^{\mathrm{3}} \:+\mathrm{3}{x}^{\mathrm{2}} \frac{\mathrm{1}}{{x}}\:+\mathrm{3}{x}\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\:=\mathrm{5}^{\mathrm{3}} \:\Rightarrow \\ $$$${x}^{\mathrm{3}} \:+{x}^{−\mathrm{3}} \:+\mathrm{3}\left({x}+\frac{\mathrm{1}}{{x}}\right)\:=\mathrm{5}^{\mathrm{3}} \:\Rightarrow{x}^{\mathrm{3}} \:+{x}^{−\mathrm{3}} \:=\mathrm{5}^{\mathrm{3}} −\mathrm{3}×\mathrm{5}\:\:\:\:\:{we}\:{have}\:\:{C}_{\mathrm{5}} ^{\mathrm{2}} \:=\frac{\mathrm{5}!}{\mathrm{2}!\mathrm{3}!} \\ $$$$=\frac{\mathrm{5}.\mathrm{4}}{\mathrm{2}}\:=\mathrm{10}\:\Rightarrow{x}^{\mathrm{5}} \:+{x}^{−\mathrm{5}} \:=\mathrm{5}^{\mathrm{5}} \:−\mathrm{5}\left(\mathrm{5}^{\mathrm{3}} −\mathrm{15}\right)−\mathrm{10}×\mathrm{5} \\ $$$$=\mathrm{5}^{\mathrm{5}} −\mathrm{5}^{\mathrm{4}} \:+\mathrm{5}×\mathrm{15}\:−\mathrm{50}\:\:=\:\mathrm{5}^{\mathrm{4}} ×\mathrm{4}\:\:\:+\mathrm{25}\:\:\:=\mathrm{4}\:\:×\mathrm{25}^{\mathrm{2}} \:+\mathrm{25}\:\:=\mathrm{4}\left(\frac{\mathrm{100}}{\mathrm{4}}\right)^{\mathrm{2}} \:+\mathrm{25} \\ $$$$=\frac{\mathrm{10}^{\mathrm{4}} }{\mathrm{4}}\:+\mathrm{25}\:\:\:=\frac{\mathrm{10000}}{\mathrm{4}}\:+\mathrm{25}\:=\mathrm{2500}\:+\mathrm{25}\:=\mathrm{2525}\:. \\ $$
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]
$${x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }= \\ $$$$=\left({x}+\frac{\mathrm{1}}{{x}}\right)\left({x}^{\mathrm{4}} −{x}^{\mathrm{3}} ×\frac{\mathrm{1}}{{x}}+{x}^{\mathrm{2}} ×\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−{x}×\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\right) \\ $$$$=\mathrm{5}\left({x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\right) \\ $$$$=\mathrm{5}\left({x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }+\mathrm{2}{x}^{\mathrm{2}} ×\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−{x}^{\mathrm{2}} −\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\mathrm{2}{x}×\frac{\mathrm{1}}{{x}}+\mathrm{1}−\mathrm{2}+\mathrm{2}\right) \\ $$$$=\mathrm{5}\left[\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{2}} −\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{1}\right] \\ $$$$=\mathrm{5}\left[\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{2}{x}×\frac{\mathrm{1}}{{x}}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{5}^{\mathrm{2}} +\mathrm{1}\right] \\ $$$$=\mathrm{5}\left[\left\{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}\right\}^{\mathrm{2}} −\mathrm{5}^{\mathrm{2}} +\mathrm{1}\right] \\ $$$$=\mathrm{5}\left[\left\{\mathrm{5}^{\mathrm{2}} −\mathrm{2}\right\}^{\mathrm{2}} −\mathrm{5}^{\mathrm{2}} +\mathrm{1}\right] \\ $$$$=\mathrm{5}\left[\mathrm{23}^{\mathrm{2}} −\mathrm{5}^{\mathrm{2}} +\mathrm{1}\right] \\ $$$$=\mathrm{5}\left[\mathrm{28}×\mathrm{18}+\mathrm{1}\right] \\ $$$$=\mathrm{5}×\mathrm{505} \\ $$$$=\mathrm{2525} \\ $$$${in}\:{general}: \\ $$$$={a}\left[\left({a}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} +\mathrm{1}\right] \\ $$$$={a}\left[\left\{{a}\left({a}+\mathrm{1}\right)−\mathrm{2}\right\}\left\{\left({a}−\mathrm{1}\right){a}−\mathrm{2}\right\}+\mathrm{1}\right] \\ $$
Commented by Akashuac last updated on 17/Aug/18
Thanks you Boss
$$\mathrm{Thanks}\:\mathrm{you}\:\mathrm{Boss} \\ $$
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
$${x}+\frac{\mathrm{1}}{{x}}={a} \\ $$$${x}^{\mathrm{2}} −{ax}+\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{{a}\pm\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{x}}=\frac{\mathrm{2}}{{a}\pm\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}}=\frac{\mathrm{2}\left({a}\mp\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}\right)}{{a}^{\mathrm{2}} −\left({a}^{\mathrm{2}} −\mathrm{4}\right)}=\frac{{a}\mp\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{write}\:{x}\:\mathrm{as}\:{p}\pm\sqrt{{q}}\:\mathrm{and}\:\frac{\mathrm{1}}{{x}}\:\mathrm{as}\:{p}\mp\sqrt{{q}} \\ $$$${p}\in\mathbb{Q}\:\Rightarrow\:{x}^{{n}} +\frac{\mathrm{1}}{{x}^{{n}} }=\left[\left({p}+\sqrt{{q}}\right)^{{n}} +\left({p}−\sqrt{{q}}\right)^{{n}} \right]\in\mathbb{Q}\:\forall{n}\in\mathbb{Z} \\ $$$$\mathrm{we}\:\mathrm{only}\:\mathrm{need}\:\mathrm{to}\:\mathrm{sum}\:\mathrm{up}\:\mathrm{the}\:{p}^{{i}} \left(\sqrt{{q}}\right)^{{j}} \:\mathrm{with}\:{j}=\mathrm{2}{k} \\ $$$${n}=\mathrm{5}\:\Rightarrow\:{x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }= \\ $$$$=\mathrm{2}\left({p}^{\mathrm{5}} +\mathrm{10}{p}^{\mathrm{3}} {q}+\mathrm{5}{pq}^{\mathrm{2}} \right)= \\ $$$$\:\:\:\:\:\left[{p}=\frac{{a}}{\mathrm{2}}\:\:{q}=\frac{{a}^{\mathrm{2}} −\mathrm{4}}{\mathrm{4}}\right] \\ $$$$={a}\left({a}^{\mathrm{4}} −\mathrm{5}{a}^{\mathrm{2}} +\mathrm{5}\right)= \\ $$$$\:\:\:\:\:\left[{a}=\mathrm{5}\right] \\ $$$$=\mathrm{2525} \\ $$
Commented by Akashuac last updated on 17/Aug/18
thanks boss
$$\mathrm{thanks}\:\mathrm{boss} \\ $$
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
$$\mathrm{by}\:\mathrm{using} \\ $$$$\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{5}} =\mathrm{5}^{\mathrm{5}} \\ $$$$\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{3}} =\mathrm{5}^{\mathrm{3}} \:\mathrm{we}\:\mathrm{find} \\ $$$$\mathrm{x}^{\mathrm{5}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{5}} }=\mathrm{5}^{\mathrm{5}} −\mathrm{10}×\mathrm{5}− \\ $$$$\:\:\:\:−\mathrm{5}\left(\mathrm{5}^{\mathrm{3}} −\mathrm{3}×\mathrm{5}\right) \\ $$$$=\mathrm{2525} \\ $$
Commented by Akashuac last updated on 17/Aug/18
thanks boss
$$\mathrm{thanks}\:\mathrm{boss} \\ $$
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
$$\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} =\mathrm{25} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{23}………\mathrm{1} \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\mathrm{3}\left({x}+\frac{\mathrm{1}}{{x}}\right)=\mathrm{125} \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\mathrm{125}−\mathrm{15}=\mathrm{110}……….\mathrm{2} \\ $$$$\therefore\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\left({x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)=\mathrm{23}×\mathrm{110} \\ $$$${x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }+{x}+\frac{\mathrm{1}}{{x}}=\mathrm{2530} \\ $$$$\therefore{x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }=\mathrm{2530}−\mathrm{5}=\mathrm{2525} \\ $$
Commented by Akashuac last updated on 17/Aug/18
thanks boss
$$\mathrm{thanks}\:\mathrm{boss} \\ $$

Pin

Leave a Reply

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