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2015-04-14T14:18:57+05:30
\sum\limits_{i=0}^nx^n = 1\\\dfrac{x^{n+1}-1}{x-1}=1\\x^{n+1}-1 = x-1\\x^{n+1}- = x\\x(x^n-1)=0

use zero product property and solve x
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2015-04-14T18:28:58+05:30

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X° + x¹ + x² + x³ + x⁴ + ..... + x^n = 1

this is a Geometric series with starting term x° = 1, and ratio = x.  there are n terms.  The sum is :

S = 1 (x^n+1  - 1) / (x - 1)
   = 1

=>  x^n+1 - 1 = x - 1
=>  x^n+1 = x
  => x * x^n = x          => x ( x^n -1 ) = 0

     => x = 0 or,    x^n = 1
  So the solutions are  :    x = 0  or the  n th roots of 1, but x ≠ 1.

  For example if  n =3,  then x = 0 or the cube roots of 1,  but x ≠ 1.
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1 + x + x² + x³ + ... + x^n = 1
x ( 1 + x + x² + x³... + x^n-1 ) = 0    ----  (1)
  = >    x = 0  or  (x^n -1 ) / (x -1) = 0     
  =>    x = 0  or  x^n = 1          and  x ≠ 1
   =>   x = n th roots of 1.

 for example  x = ω or ω² for n = 3.   
     as 1 + ω + ω² = 0, the above equation (1) is satisfied.
   so then roots will be  0 , ω or ω².

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