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Question Number 151699 by iloveisrael last updated on 22/Aug/21
Σ_(x^3 +2022x−2021=0) (((1+x)/(1−x))) =?
$$\:\:\:\:\underset{{x}^{\mathrm{3}} +\mathrm{2022}{x}−\mathrm{2021}=\mathrm{0}} {\sum}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)\:=? \\ $$
Answered by mr W last updated on 22/Aug/21
x^3 +2022x−2021=0 (x−1+1)^3 +2022(x−1+1)−2021=0 let x−1=t (t+1)^3 +2022(t+1)−2021=0 t^3 +3t^2 +3t+1+2022t+2022−2021=0 t^3 +3t^2 +2025t+2=0 (2/t^3 )+((2025)/t^2 )+(3/t)+1=0 ⇒Σ(1/t)=−((2025)/2) ((1+x)/(1−x))=−((x−1+2)/(x−1))=−1−(2/(x−1))=−1−(2/t) Σ((1+x)/(1−x))=Σ(−1−(2/t))=−3−2Σ(1/t) =−3−2×(−((2025)/2)) =2022
$${x}^{\mathrm{3}} +\mathrm{2022}{x}−\mathrm{2021}=\mathrm{0} \\ $$$$\left({x}−\mathrm{1}+\mathrm{1}\right)^{\mathrm{3}} +\mathrm{2022}\left({x}−\mathrm{1}+\mathrm{1}\right)−\mathrm{2021}=\mathrm{0} \\ $$$${let}\:{x}−\mathrm{1}={t} \\ $$$$\left({t}+\mathrm{1}\right)^{\mathrm{3}} +\mathrm{2022}\left({t}+\mathrm{1}\right)−\mathrm{2021}=\mathrm{0} \\ $$$${t}^{\mathrm{3}} +\mathrm{3}{t}^{\mathrm{2}} +\mathrm{3}{t}+\mathrm{1}+\mathrm{2022}{t}+\mathrm{2022}−\mathrm{2021}=\mathrm{0} \\ $$$${t}^{\mathrm{3}} +\mathrm{3}{t}^{\mathrm{2}} +\mathrm{2025}{t}+\mathrm{2}=\mathrm{0} \\ $$$$\frac{\mathrm{2}}{{t}^{\mathrm{3}} }+\frac{\mathrm{2025}}{{t}^{\mathrm{2}} }+\frac{\mathrm{3}}{{t}}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\Sigma\frac{\mathrm{1}}{{t}}=−\frac{\mathrm{2025}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}=−\frac{{x}−\mathrm{1}+\mathrm{2}}{{x}−\mathrm{1}}=−\mathrm{1}−\frac{\mathrm{2}}{{x}−\mathrm{1}}=−\mathrm{1}−\frac{\mathrm{2}}{{t}} \\ $$$$\Sigma\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}=\Sigma\left(−\mathrm{1}−\frac{\mathrm{2}}{{t}}\right)=−\mathrm{3}−\mathrm{2}\Sigma\frac{\mathrm{1}}{{t}} \\ $$$$=−\mathrm{3}−\mathrm{2}×\left(−\frac{\mathrm{2025}}{\mathrm{2}}\right) \\ $$$$=\mathrm{2022} \\ $$
Commented by iloveisrael last updated on 22/Aug/21
thank you
$${thank}\:{you} \\ $$
Answered by aleks041103 last updated on 22/Aug/21
((1+x)/(1−x))=((1+1−(1−x))/(1−x))=(2/(1−x))−1 x^3 +2022x−2021=0 has 3 roots by the fundamental theorem of algebra. ⇒Σ_(i=1) ^3 ((2/(1−x_i ))−1)=2(Σ_(i=1) ^3 (1/(1−x_i )))−3 Σ_(i=1) ^3 (1/(1−x_i ))=(((1−x_2 )(1−x_3 )+(1−x_1 )(1−x_3 )+(1−x_1 )(1−x_2 ))/((1−x_1 )(1−x_2 )(1−x_3 )))= =((3−2(x_1 +x_2 +x_3 )+(x_1 x_2 +x_2 x_3 +x_1 x_3 ))/((1−x_1 )(1−x_2 )(1−x_3 ))) Since x_(1,2,3) are roots of the cubic polynomial then x^3 +2022x−2021=(x−x_1 )(x−x_2 )(x−x_3 ) Then (1−x_1 )(1−x_2 )(1−x_3 )=1^3 +2022.1−2021=2 Also by Vieta′s formulas x_1 +x_2 +x_3 =0 x_1 x_2 +x_2 x_3 +x_1 x_3 =2022 Therefore Σ_(i=1) ^3 (1/(1−x_i ))=((3−2(0)+2022)/2)=((2025)/2) Then: Σ_(x^3 +2022x−2021=0) (((1+x)/(1−x))) =2022
$$\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}=\frac{\mathrm{1}+\mathrm{1}−\left(\mathrm{1}−{x}\right)}{\mathrm{1}−{x}}=\frac{\mathrm{2}}{\mathrm{1}−{x}}−\mathrm{1} \\ $$$${x}^{\mathrm{3}} +\mathrm{2022}{x}−\mathrm{2021}=\mathrm{0}\:{has}\:\mathrm{3}\:{roots}\:{by} \\ $$$${the}\:{fundamental}\:{theorem}\:{of}\:{algebra}. \\ $$$$\Rightarrow\underset{{i}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}\left(\frac{\mathrm{2}}{\mathrm{1}−{x}_{{i}} }−\mathrm{1}\right)=\mathrm{2}\left(\underset{{i}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}\frac{\mathrm{1}}{\mathrm{1}−{x}_{{i}} }\right)−\mathrm{3} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}\frac{\mathrm{1}}{\mathrm{1}−{x}_{{i}} }=\frac{\left(\mathrm{1}−{x}_{\mathrm{2}} \right)\left(\mathrm{1}−{x}_{\mathrm{3}} \right)+\left(\mathrm{1}−{x}_{\mathrm{1}} \right)\left(\mathrm{1}−{x}_{\mathrm{3}} \right)+\left(\mathrm{1}−{x}_{\mathrm{1}} \right)\left(\mathrm{1}−{x}_{\mathrm{2}} \right)}{\left(\mathrm{1}−{x}_{\mathrm{1}} \right)\left(\mathrm{1}−{x}_{\mathrm{2}} \right)\left(\mathrm{1}−{x}_{\mathrm{3}} \right)}= \\ $$$$=\frac{\mathrm{3}−\mathrm{2}\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} \right)+\left({x}_{\mathrm{1}} {x}_{\mathrm{2}} +{x}_{\mathrm{2}} {x}_{\mathrm{3}} +{x}_{\mathrm{1}} {x}_{\mathrm{3}} \right)}{\left(\mathrm{1}−{x}_{\mathrm{1}} \right)\left(\mathrm{1}−{x}_{\mathrm{2}} \right)\left(\mathrm{1}−{x}_{\mathrm{3}} \right)} \\ $$$${Since}\:{x}_{\mathrm{1},\mathrm{2},\mathrm{3}} \:{are}\:{roots}\:{of}\:{the}\:{cubic}\:{polynomial} \\ $$$${then}\: \\ $$$${x}^{\mathrm{3}} +\mathrm{2022}{x}−\mathrm{2021}=\left({x}−{x}_{\mathrm{1}} \right)\left({x}−{x}_{\mathrm{2}} \right)\left({x}−{x}_{\mathrm{3}} \right) \\ $$$${Then} \\ $$$$\left(\mathrm{1}−{x}_{\mathrm{1}} \right)\left(\mathrm{1}−{x}_{\mathrm{2}} \right)\left(\mathrm{1}−{x}_{\mathrm{3}} \right)=\mathrm{1}^{\mathrm{3}} +\mathrm{2022}.\mathrm{1}−\mathrm{2021}=\mathrm{2} \\ $$$${Also}\:{by}\:{Vieta}'{s}\:{formulas} \\ $$$${x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} =\mathrm{0} \\ $$$${x}_{\mathrm{1}} {x}_{\mathrm{2}} +{x}_{\mathrm{2}} {x}_{\mathrm{3}} +{x}_{\mathrm{1}} {x}_{\mathrm{3}} =\mathrm{2022} \\ $$$${Therefore} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}\frac{\mathrm{1}}{\mathrm{1}−{x}_{{i}} }=\frac{\mathrm{3}−\mathrm{2}\left(\mathrm{0}\right)+\mathrm{2022}}{\mathrm{2}}=\frac{\mathrm{2025}}{\mathrm{2}} \\ $$$${Then}: \\ $$$$\:\underset{{x}^{\mathrm{3}} +\mathrm{2022}{x}−\mathrm{2021}=\mathrm{0}} {\sum}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)\:=\mathrm{2022} \\ $$
Commented by iloveisrael last updated on 22/Aug/21
thank you
$${thank}\:{you} \\ $$

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