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Question Number 157695 by amin96 last updated on 26/Oct/21
Answered by Rasheed.Sindhi last updated on 26/Oct/21
x^(16) +(1/x^(16) )=(x^8 +(1/x^8 ))^2 −2=2207 x^8 +(1/x^8 )=(√(2209)) =47 x^8 +(1/x^8 )=(x^4 +(1/x^4 ))^2 −2=47 x^4 +(1/x^4 )=(√(49)) =7 x^4 +(1/x^4 )=(x^2 +(1/x^2 ))^2 −2=7 x^2 +(1/x^2 )=(√9) =3 x^2 +(1/x^2 )=(x+(1/x))^2 −2=3 x+(1/x)=(√5) ....... .....
$${x}^{\mathrm{16}} +\frac{\mathrm{1}}{{x}^{\mathrm{16}} }=\left({x}^{\mathrm{8}} +\frac{\mathrm{1}}{{x}^{\mathrm{8}} }\right)^{\mathrm{2}} −\mathrm{2}=\mathrm{2207} \\ $$$${x}^{\mathrm{8}} +\frac{\mathrm{1}}{{x}^{\mathrm{8}} }=\sqrt{\mathrm{2209}}\:=\mathrm{47} \\ $$$${x}^{\mathrm{8}} +\frac{\mathrm{1}}{{x}^{\mathrm{8}} }=\left({x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\right)^{\mathrm{2}} −\mathrm{2}=\mathrm{47} \\ $$$${x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }=\sqrt{\mathrm{49}}\:=\mathrm{7} \\ $$$${x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }=\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{2}} −\mathrm{2}=\mathrm{7} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\sqrt{\mathrm{9}}\:=\mathrm{3} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}=\mathrm{3} \\ $$$${x}+\frac{\mathrm{1}}{{x}}=\sqrt{\mathrm{5}} \\ $$$$……. \\ $$$$….. \\ $$
Commented by amin96 last updated on 26/Oct/21
I know this sir . how will it be after ?
$$ \\ $$I know this sir . how will it be after ?
Commented by Rasheed.Sindhi last updated on 26/Oct/21
Very hard to continue further sir.I have done only easy part. :)
Answered by mr W last updated on 26/Oct/21
(x^8 +(1/x^8 ))^2 =2209=47^2 x^8 +(1/x^8 )=47 (x^4 +(1/x^4 ))^2 =49 x^4 +(1/x^4 )=7 (x^2 +(1/x^2 ))^2 =9 x^2 +(1/x^2 )=3 let t=x^2 t+(1/t)=3 t^2 −3t+1=0 p_n =t^n +(1/t^n )=x^(2n) +(1/x^(2n) ) p_(n+1) =3p_n −p_(n−1) r^2 −3r+1=0 r_(1,2) =((3±(√5))/2) p_n =r_1 ^n +r_2 ^n =(((3+(√5))/2))^n +(((3−(√5))/2))^n a_n =(x^n +(1/x^n ))^2 =x^(2n) +(1/x^(2n) )+2=p_n +2 Σ_(n=1) ^(m=2021) a_n =Σ_(n=1) ^m (r_1 ^n +r_2 ^n +2) =((r_1 (r_1 ^m −1))/(r_1 −1))+((r_2 (r_2 ^m −1))/(r_2 −1))+2m =(((((3+(√5))/2))[(((3+(√5))/2))^m −1])/(((3+(√5))/2)−1))+(((((3−(√5))/2))[(((3−(√5))/2))^m −1])/(((3−(√5))/2)−1))+2m =(((3+(√5))[(((3+(√5))/2))^m −1])/( (√5)+1))−(((3−(√5))[(((3−(√5))/2))^m −1])/( (√5)−1))+2m =((((√5)+1)/2))(((3+(√5))/2))^m −((((√5)−1)/2))(((3−(√5))/2))^m +2m−1
$$\left({x}^{\mathrm{8}} +\frac{\mathrm{1}}{{x}^{\mathrm{8}} }\right)^{\mathrm{2}} =\mathrm{2209}=\mathrm{47}^{\mathrm{2}} \\ $$$${x}^{\mathrm{8}} +\frac{\mathrm{1}}{{x}^{\mathrm{8}} }=\mathrm{47} \\ $$$$\left({x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\right)^{\mathrm{2}} =\mathrm{49} \\ $$$${x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }=\mathrm{7} \\ $$$$\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{2}} =\mathrm{9} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{3} \\ $$$${let}\:{t}={x}^{\mathrm{2}} \\ $$$${t}+\frac{\mathrm{1}}{{t}}=\mathrm{3} \\ $$$${t}^{\mathrm{2}} −\mathrm{3}{t}+\mathrm{1}=\mathrm{0} \\ $$$${p}_{{n}} ={t}^{{n}} +\frac{\mathrm{1}}{{t}^{{n}} }={x}^{\mathrm{2}{n}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}{n}} } \\ $$$${p}_{{n}+\mathrm{1}} =\mathrm{3}{p}_{{n}} −{p}_{{n}−\mathrm{1}} \\ $$$${r}^{\mathrm{2}} −\mathrm{3}{r}+\mathrm{1}=\mathrm{0} \\ $$$${r}_{\mathrm{1},\mathrm{2}} =\frac{\mathrm{3}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${p}_{{n}} ={r}_{\mathrm{1}} ^{{n}} +{r}_{\mathrm{2}} ^{{n}} =\left(\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} +\left(\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \\ $$$${a}_{{n}} =\left({x}^{{n}} +\frac{\mathrm{1}}{{x}^{{n}} }\right)^{\mathrm{2}} ={x}^{\mathrm{2}{n}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}{n}} }+\mathrm{2}={p}_{{n}} +\mathrm{2} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{{m}=\mathrm{2021}} {\sum}}{a}_{{n}} =\underset{{n}=\mathrm{1}} {\overset{{m}} {\sum}}\left({r}_{\mathrm{1}} ^{{n}} +{r}_{\mathrm{2}} ^{{n}} +\mathrm{2}\right) \\ $$$$=\frac{{r}_{\mathrm{1}} \left({r}_{\mathrm{1}} ^{{m}} −\mathrm{1}\right)}{{r}_{\mathrm{1}} −\mathrm{1}}+\frac{{r}_{\mathrm{2}} \left({r}_{\mathrm{2}} ^{{m}} −\mathrm{1}\right)}{{r}_{\mathrm{2}} −\mathrm{1}}+\mathrm{2}{m} \\ $$$$=\frac{\left(\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\left[\left(\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{m}} −\mathrm{1}\right]}{\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}−\mathrm{1}}+\frac{\left(\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\left[\left(\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{m}} −\mathrm{1}\right]}{\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}−\mathrm{1}}+\mathrm{2}{m} \\ $$$$=\frac{\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)\left[\left(\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{m}} −\mathrm{1}\right]}{\:\sqrt{\mathrm{5}}+\mathrm{1}}−\frac{\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)\left[\left(\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{m}} −\mathrm{1}\right]}{\:\sqrt{\mathrm{5}}−\mathrm{1}}+\mathrm{2}{m} \\ $$$$=\left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}\right)\left(\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{m}} −\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right)\left(\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{m}} +\mathrm{2}{m}−\mathrm{1} \\ $$
Answered by mindispower last updated on 26/Oct/21
U_n =x^n +(1/x^n ),u_0 =2,u_1 =(√5) U_(n+1) =(√5)U_n −U_(n−1) X^2 −(√5)X+1=0 ⇒X∈{(((√5)−1)/2),((1+(√5))/2)} U_n =a(((1+(√5))/2))^n +b((((√5)−1)/2))^n a+b=2,(1/2)(a−b)+((√5)/2)(a+b)=(√5) a=b⇒a=b=1 U_n =(((1+(√5))/2))^n +((((√5)−1)/2))^n we want Σ_(n=1) ^(2207) (U_(2n) +2) 2.2207+Σ_(n=1) ^(2207) U_(2n) =(((1+(√5))/2))^2 .((1−(((1+(√5))/2))^(4414) )/(1−(((1+(√5))/2))^2 ))+((((√5)−1)/2))^2 .((1−((((√5)−1)/2))^(4414) )/(1−((((√5)−1)/2))^2 ))+4414
$${U}_{{n}} ={x}^{{n}} +\frac{\mathrm{1}}{{x}^{{n}} },{u}_{\mathrm{0}} =\mathrm{2},{u}_{\mathrm{1}} =\sqrt{\mathrm{5}} \\ $$$${U}_{{n}+\mathrm{1}} =\sqrt{\mathrm{5}}{U}_{{n}} −{U}_{{n}−\mathrm{1}} \\ $$$${X}^{\mathrm{2}} −\sqrt{\mathrm{5}}{X}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{X}\in\left\{\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right\} \\ $$$${U}_{{n}} ={a}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} +{b}\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right)^{{n}} \\ $$$${a}+{b}=\mathrm{2},\frac{\mathrm{1}}{\mathrm{2}}\left({a}−{b}\right)+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\left({a}+{b}\right)=\sqrt{\mathrm{5}} \\ $$$${a}={b}\Rightarrow{a}={b}=\mathrm{1} \\ $$$${U}_{{n}} =\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} +\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right)^{{n}} \\ $$$${we}\:{want}\: \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{2207}} {\sum}}\left({U}_{\mathrm{2}{n}} +\mathrm{2}\right) \\ $$$$\mathrm{2}.\mathrm{2207}+\underset{{n}=\mathrm{1}} {\overset{\mathrm{2207}} {\sum}}{U}_{\mathrm{2}{n}} =\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} .\frac{\mathrm{1}−\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{4414}} }{\mathrm{1}−\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} }+\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} .\frac{\mathrm{1}−\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{4414}} }{\mathrm{1}−\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }+\mathrm{4414} \\ $$$$ \\ $$
Answered by Raxreedoroid last updated on 28/Oct/21
x^(16) +(1/x^(16) )=(x^8 +(1/x^8 ))^2 −2=2207 x^8 +(1/x^8 )=(√(2209)) =47 x^8 +(1/x^8 )=(x^4 +(1/x^4 ))^2 −2=47 x^4 +(1/x^4 )=(√(49)) =7 x^4 +(1/x^4 )=(x^2 +(1/x^2 ))^2 −2=7 x^2 +(1/x^2 )=(√9) =3 x^2 +(1/x^2 )=(x+(1/x))^2 −2=3 x+(1/x)=(√5) x^2 −x(√5)+1=0 x=((±1+(√5))/2)=ϕ^(±1) ϕ^n =ϕ^(n−2) +ϕ^(n−1) ϕ^n =ϕ^(n+2) −ϕ^(n+1) S=Σ_(n=1) ^(2021) (ϕ^n +ϕ^(−n) )^2 S=Σ_(n=1) ^(2021) (ϕ^(2n) +2+ϕ^(−2n) ) S=(((ϕ^2 )((ϕ^2 )^(2021) −1))/(ϕ^2 −1))+2(2021)+(((ϕ^(−2) )((ϕ^(−2) )^(2021) −1))/(ϕ^(−2) −1)) S=(((ϕ^2 )((ϕ)^(4042) −1))/ϕ)+4042−(((ϕ^(−2) )((ϕ)^(−4042) −1))/ϕ^(−1) ) S=ϕ^(4043) +4041−(ϕ^(−1) )^(4043)
$$ \\ $$$${x}^{\mathrm{16}} +\frac{\mathrm{1}}{{x}^{\mathrm{16}} }=\left({x}^{\mathrm{8}} +\frac{\mathrm{1}}{{x}^{\mathrm{8}} }\right)^{\mathrm{2}} −\mathrm{2}=\mathrm{2207} \\ $$$${x}^{\mathrm{8}} +\frac{\mathrm{1}}{{x}^{\mathrm{8}} }=\sqrt{\mathrm{2209}}\:=\mathrm{47} \\ $$$${x}^{\mathrm{8}} +\frac{\mathrm{1}}{{x}^{\mathrm{8}} }=\left({x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\right)^{\mathrm{2}} −\mathrm{2}=\mathrm{47} \\ $$$${x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }=\sqrt{\mathrm{49}}\:=\mathrm{7} \\ $$$${x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }=\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{2}} −\mathrm{2}=\mathrm{7} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\sqrt{\mathrm{9}}\:=\mathrm{3} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}=\mathrm{3} \\ $$$${x}+\frac{\mathrm{1}}{{x}}=\sqrt{\mathrm{5}} \\ $$$${x}^{\mathrm{2}} −{x}\sqrt{\mathrm{5}}+\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{\pm\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}=\varphi^{\pm\mathrm{1}} \\ $$$$\varphi^{{n}} =\varphi^{{n}−\mathrm{2}} +\varphi^{{n}−\mathrm{1}} \\ $$$$\varphi^{{n}} =\varphi^{{n}+\mathrm{2}} −\varphi^{{n}+\mathrm{1}} \\ $$$${S}=\underset{{n}=\mathrm{1}} {\overset{\mathrm{2021}} {\sum}}\left(\varphi^{{n}} +\varphi^{−{n}} \right)^{\mathrm{2}} \\ $$$${S}=\underset{{n}=\mathrm{1}} {\overset{\mathrm{2021}} {\sum}}\left(\varphi^{\mathrm{2}{n}} +\mathrm{2}+\varphi^{−\mathrm{2}{n}} \right) \\ $$$${S}=\frac{\left(\varphi^{\mathrm{2}} \right)\left(\left(\varphi^{\mathrm{2}} \right)^{\mathrm{2021}} −\mathrm{1}\right)}{\varphi^{\mathrm{2}} −\mathrm{1}}+\mathrm{2}\left(\mathrm{2021}\right)+\frac{\left(\varphi^{−\mathrm{2}} \right)\left(\left(\varphi^{−\mathrm{2}} \right)^{\mathrm{2021}} −\mathrm{1}\right)}{\varphi^{−\mathrm{2}} −\mathrm{1}} \\ $$$${S}=\frac{\left(\varphi^{\mathrm{2}} \right)\left(\left(\varphi\right)^{\mathrm{4042}} −\mathrm{1}\right)}{\varphi}+\mathrm{4042}−\frac{\left(\varphi^{−\mathrm{2}} \right)\left(\left(\varphi\right)^{−\mathrm{4042}} −\mathrm{1}\right)}{\varphi^{−\mathrm{1}} } \\ $$$${S}=\varphi^{\mathrm{4043}} +\mathrm{4041}−\left(\varphi^{−\mathrm{1}} \right)^{\mathrm{4043}} \\ $$

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