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Question Number 55992 by Mikael_Marshall last updated on 07/Mar/19

1 + x + x^2 + . . . + x^(99) =0 need an explanation.

1+x+x2+...+x99=0needanexplanation.

Commented by mr W last updated on 07/Mar/19

if x=−1: 1 + x + x^2 +x^3 + . . .+x^(98) + x^(99) =(1 + x) + (x^2 +x^3 )+ . . .+(x^(98) + x^(99) ) =(1−1)+(1−1)+...+(1−1) =0 or if x=−1: 1 + x + x^2 + . . . + x^(99) =((1−x^(100) )/(1−x))=((1−1)/2)=0

ifx=−1:1+x+x2+x3+...+x98+x99=(1+x)+(x2+x3)+...+(x98+x99)=(1−1)+(1−1)+...+(1−1)=0orifx=−1:1+x+x2+...+x99=1−x1001−x=1−12=0

Commented by maxmathsup by imad last updated on 07/Mar/19

if x from C (e) ⇔ ((1−x^(100) )/(1−x)) =0 with x≠1 ⇒x^(100) =1 =e^(i2kπ) ⇒ x_k =e^((ikπ)/(50)) and k∈[[1,99]] are roots of this equation.

ifxfromC(e)⇔1−x1001−x=0withx≠1⇒x100=1=ei2kπ⇒xk=eikπ50andk∈[[1,99]]arerootsofthisequation.

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