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Question Number 12797 by tawa last updated on 01/May/17

1 + x + x^2 + ... x^(49) = (1/2)(x^(49) − (1/x)) Find the value of x

$$\mathrm{1}\:+\:\mathrm{x}\:+\:\mathrm{x}^{\mathrm{2}} \:+\:...\:\mathrm{x}^{\mathrm{49}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{x}^{\mathrm{49}} \:−\:\frac{\mathrm{1}}{\mathrm{x}}\right) \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x} \\ $$

Commented by prakash jain last updated on 01/May/17

((x^(50) −1)/(x−1))=(1/2)(((x^(50) −1)/x)) x≠1,x≠0 (x^(50) −1)2x=(x−1)(x^(50) −1) (x^(50) −1)(x+1)=0 x=−1 x^(50) =1⇒x=cos ((2kπ)/(50))+isin ((2kπ)/(50)) k=1,2,...,49 k=0⇒x=1, invalid since x≠1

$$\frac{{x}^{\mathrm{50}} −\mathrm{1}}{{x}−\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{x}^{\mathrm{50}} −\mathrm{1}}{{x}}\right) \\ $$$${x}\neq\mathrm{1},{x}\neq\mathrm{0} \\ $$$$\left({x}^{\mathrm{50}} −\mathrm{1}\right)\mathrm{2}{x}=\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{50}} −\mathrm{1}\right) \\ $$$$\left({x}^{\mathrm{50}} −\mathrm{1}\right)\left({x}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}=−\mathrm{1} \\ $$$${x}^{\mathrm{50}} =\mathrm{1}\Rightarrow{x}=\mathrm{cos}\:\frac{\mathrm{2}{k}\pi}{\mathrm{50}}+{i}\mathrm{sin}\:\frac{\mathrm{2}{k}\pi}{\mathrm{50}}\: \\ $$$${k}=\mathrm{1},\mathrm{2},...,\mathrm{49} \\ $$$${k}=\mathrm{0}\Rightarrow{x}=\mathrm{1},\:{invalid}\:{since}\:{x}\neq\mathrm{1} \\ $$

Commented by tawa last updated on 01/May/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 01/May/17

2(x^(50) −1)=(x−1)(x^(50) −1)/x⇒ 2x=x−1⇒x=−1 .■ if :x^(50) −1=0⇒ there is no solotion for x.

$$\mathrm{2}\left({x}^{\mathrm{50}} −\mathrm{1}\right)=\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{50}} −\mathrm{1}\right)/{x}\Rightarrow \\ $$$$\mathrm{2}{x}={x}−\mathrm{1}\Rightarrow{x}=−\mathrm{1}\:\:.\blacksquare \\ $$$${if}\::{x}^{\mathrm{50}} −\mathrm{1}=\mathrm{0}\Rightarrow\:{there}\:{is}\:{no}\:{solotion}\:{for}\:{x}. \\ $$

Commented by tawa last updated on 01/May/17

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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