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Question Number 85832 by jagoll last updated on 25/Mar/20
(x+x^(−1) )^2 +(x^2 +x^(−2) )^2 +(x^3 +x^(−3) )^2 + ... + (x^(10) +x^(−10) )^2 =
$$\left(\mathrm{x}+\mathrm{x}^{−\mathrm{1}} \right)^{\mathrm{2}} +\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}^{−\mathrm{2}} \right)^{\mathrm{2}} +\left(\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{−\mathrm{3}} \right)^{\mathrm{2}} \\ $$$$+\:…\:+\:\left(\mathrm{x}^{\mathrm{10}} +\mathrm{x}^{−\mathrm{10}} \right)^{\mathrm{2}} \:=\: \\ $$
Commented by mathmax by abdo last updated on 25/Mar/20
let A_n (x)=(x+x^(−1) )^2 +(x^2 +x^(−2) )^2 +(x^3 +x^(−3) )^2 +...+(x^n +x^(−n) )^2 ⇒A_n =x^2 +(1/x^2 ) +2 +x^4 +(1/x^4 ) +2 +....x^(2n) +(1/x^(2n) ) +2 =x^2 +x^4 +....+x^(2n) +((1/x))^2 +((1/x))^4 +....+((1/x))^(2n) +2n =w(x)+w((1/x)) +2n with w(x)=x^2 +x^4 +...x^(2n) =x^2 (1+x^2 +...+(x^2 )^(n−1) ) =x^2 ×((1−(x^2 )^n )/(1−x^2 )) =(x^2 /(1−x^2 ))(1−x^(2n) ) w((1/x)) =(1/(x^2 (1−(1/x^2 ))))(1−(1/x^(2n) )) =(1/x^2 )(1−(1/x^(2n) )) =((x^(2n) −1)/x^(2n+2) ) ⇒ A_n (x)=((x^2 −x^(2n+2) )/(1−x^2 )) +((x^(2n) −1)/x^(2n+2) ) +2n (x≠+^− 1) n=10 ⇒ A_(10) (x) =((x^2 −x^(22) )/(1−x^2 )) +((x^(20) −1)/x^(22) ) +20
$${let}\:{A}_{{n}} \left({x}\right)=\left({x}+{x}^{−\mathrm{1}} \right)^{\mathrm{2}} \:+\left({x}^{\mathrm{2}} \:+{x}^{−\mathrm{2}} \right)^{\mathrm{2}} \:+\left({x}^{\mathrm{3}} \:+{x}^{−\mathrm{3}} \right)^{\mathrm{2}} +…+\left({x}^{{n}} \:+{x}^{−{n}} \right)^{\mathrm{2}} \\ $$$$\Rightarrow{A}_{{n}} ={x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\mathrm{2}\:\:+{x}^{\mathrm{4}} \:+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\:+\mathrm{2}\:+….{x}^{\mathrm{2}{n}} \:+\frac{\mathrm{1}}{{x}^{\mathrm{2}{n}} }\:+\mathrm{2} \\ $$$$={x}^{\mathrm{2}} \:+{x}^{\mathrm{4}} \:+….+{x}^{\mathrm{2}{n}} \:+\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} \:+\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{4}} \:+….+\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}{n}} \:+\mathrm{2}{n} \\ $$$$={w}\left({x}\right)+{w}\left(\frac{\mathrm{1}}{{x}}\right)\:+\mathrm{2}{n}\:{with}\:{w}\left({x}\right)={x}^{\mathrm{2}} \:+{x}^{\mathrm{4}} \:+…{x}^{\mathrm{2}{n}} \\ $$$$={x}^{\mathrm{2}} \left(\mathrm{1}+{x}^{\mathrm{2}} \:+…+\left({x}^{\mathrm{2}} \right)^{{n}−\mathrm{1}} \right)\:={x}^{\mathrm{2}} ×\frac{\mathrm{1}−\left({x}^{\mathrm{2}} \right)^{{n}} }{\mathrm{1}−{x}^{\mathrm{2}} }\:=\frac{{x}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{2}} }\left(\mathrm{1}−{x}^{\mathrm{2}{n}} \right) \\ $$$${w}\left(\frac{\mathrm{1}}{{x}}\right)\:=\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}{n}} }\right)\:=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}{n}} }\right)\:=\frac{{x}^{\mathrm{2}{n}} −\mathrm{1}}{{x}^{\mathrm{2}{n}+\mathrm{2}} }\:\Rightarrow \\ $$$${A}_{{n}} \left({x}\right)=\frac{{x}^{\mathrm{2}} −{x}^{\mathrm{2}{n}+\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{2}} }\:+\frac{{x}^{\mathrm{2}{n}} −\mathrm{1}}{{x}^{\mathrm{2}{n}+\mathrm{2}} }\:+\mathrm{2}{n}\:\:\:\:\left({x}\neq\overset{−} {+}\mathrm{1}\right) \\ $$$${n}=\mathrm{10}\:\Rightarrow\:{A}_{\mathrm{10}} \left({x}\right)\:\:=\frac{{x}^{\mathrm{2}} −{x}^{\mathrm{22}} }{\mathrm{1}−{x}^{\mathrm{2}} }\:+\frac{{x}^{\mathrm{20}} −\mathrm{1}}{{x}^{\mathrm{22}} }\:+\mathrm{20} \\ $$
Answered by TANMAY PANACEA. last updated on 25/Mar/20
T_r =(x^r +x^(−r) )^2 =x^(2r) +2+(1/x^(2r) ) T_1 +T_2 +T_3 +...+T_(10) =(x^2 +x^4 +x^6 +..+x^(20) )+2×10+((1/x^2 )+(1/x^4 )+(1/x^6 )+..+(1/x^(20) )) s=((a(1−r^n ))/(1−r)) =((x^2 (1−x^(20) ))/(1−x^2 ))+20+(((1/x^2 )(1−(1/x^(20) )))/(1−(1/x^2 )))
$${T}_{{r}} =\left({x}^{{r}} +{x}^{−{r}} \right)^{\mathrm{2}} ={x}^{\mathrm{2}{r}} +\mathrm{2}+\frac{\mathrm{1}}{{x}^{\mathrm{2}{r}} } \\ $$$${T}_{\mathrm{1}} +{T}_{\mathrm{2}} +{T}_{\mathrm{3}} +…+{T}_{\mathrm{10}} \\ $$$$=\left({x}^{\mathrm{2}} +{x}^{\mathrm{4}} +{x}^{\mathrm{6}} +..+{x}^{\mathrm{20}} \right)+\mathrm{2}×\mathrm{10}+\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }+\frac{\mathrm{1}}{{x}^{\mathrm{6}} }+..+\frac{\mathrm{1}}{{x}^{\mathrm{20}} }\right) \\ $$$${s}=\frac{{a}\left(\mathrm{1}−{r}^{{n}} \right)}{\mathrm{1}−{r}} \\ $$$$=\frac{{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{20}} \right)}{\mathrm{1}−{x}^{\mathrm{2}} }+\mathrm{20}+\frac{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{20}} }\right)}{\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by jagoll last updated on 25/Mar/20
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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