Question-177442

Question Number 177442 by peter frank last updated on 05/Oct/22

Answered by mr W last updated on 05/Oct/22

x≠1 (x−1)(1+x+...+x^(2013) )=0 x^(2014) −1=0 x^(2014) =1=e^(2kπi) ⇒x=e^((2kπi)/(2014)) =e^((kπi)/(1007)) =cos ((kπ)/(1007))+i sin ((kπ)/(1007)) with k=1,2,...,2013 if x∈R: x^(2014) =1 ⇒x=−1

$${x}\neq\mathrm{1} \\ $$$$\left({x}−\mathrm{1}\right)\left(\mathrm{1}+{x}+…+{x}^{\mathrm{2013}} \right)=\mathrm{0} \\ $$$${x}^{\mathrm{2014}} −\mathrm{1}=\mathrm{0} \\ $$$${x}^{\mathrm{2014}} =\mathrm{1}={e}^{\mathrm{2}{k}\pi{i}} \\ $$$$\Rightarrow{x}={e}^{\frac{\mathrm{2}{k}\pi{i}}{\mathrm{2014}}} ={e}^{\frac{{k}\pi{i}}{\mathrm{1007}}} =\mathrm{cos}\:\frac{{k}\pi}{\mathrm{1007}}+{i}\:\mathrm{sin}\:\frac{{k}\pi}{\mathrm{1007}} \\ $$$${with}\:{k}=\mathrm{1},\mathrm{2},…,\mathrm{2013} \\ $$$$ \\ $$$${if}\:{x}\in\mathbb{R}: \\ $$$${x}^{\mathrm{2014}} =\mathrm{1}\:\Rightarrow{x}=−\mathrm{1} \\ $$

Commented by peter frank last updated on 05/Oct/22

$$\mathrm{thank}\:\mathrm{you} \\ $$

Commented by Tawa11 last updated on 05/Oct/22

$$\mathrm{Great}\:\mathrm{sir}. \\ $$

Answered by Strengthenchen last updated on 05/Oct/22

As I thought,the question could be seem as a geometric progression,deal it like this: ((1(1−x^(2014) ))/(1−x))=0 so (1−x^(2014) )=0, easy to know x=±1,1 is not fit question,−1 is finally answer

$${As}\:{I}\:{thought},{the}\:{question}\:{could}\:{be}\:{seem}\:{as}\:{a}\:{geometric}\:{progression},{deal}\:{it}\:{like}\:{this}: \\ $$$$\frac{\mathrm{1}\left(\mathrm{1}−{x}^{\mathrm{2014}} \right)}{\mathrm{1}−{x}}=\mathrm{0} \\ $$$${so}\:\left(\mathrm{1}−{x}^{\mathrm{2014}} \right)=\mathrm{0},\:{easy}\:{to}\:{know}\:{x}=\pm\mathrm{1},\mathrm{1}\:{is}\:{not}\:{fit}\:{question},−\mathrm{1}\:{is}\:{finally}\:{answer} \\ $$

Commented by mr W last updated on 06/Oct/22

as for x∈C, 1−x^(2014) =0 has totally 2014 roots.

$${as}\:{for}\:{x}\in\mathbb{C},\:\mathrm{1}−{x}^{\mathrm{2014}} =\mathrm{0}\:{has}\:{totally} \\ $$$$\mathrm{2014}\:{roots}. \\ $$

Answered by Rasheed.Sindhi last updated on 05/Oct/22

1+x+x^2 +x^3 +...+x^(2012) +x^(2013) =0 (1+x)+x^2 (1+x)+...+x^(2012) (1+x)=0 (1+x)(1+x^2 +x^4 +...+x^(2012) )=0 1+x=0⇒x=−1

$$\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…+{x}^{\mathrm{2012}} +{x}^{\mathrm{2013}} =\mathrm{0} \\ $$$$\left(\mathrm{1}+{x}\right)+{x}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)+…+{x}^{\mathrm{2012}} \left(\mathrm{1}+{x}\right)=\mathrm{0} \\ $$$$\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{4}} +…+{x}^{\mathrm{2012}} \right)=\mathrm{0} \\ $$$$\mathrm{1}+{x}=\mathrm{0}\Rightarrow{x}=−\mathrm{1} \\ $$

Commented by peter frank last updated on 05/Oct/22

$$\mathrm{thank}\:\mathrm{you} \\ $$

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