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

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

	$x \neq 1$ $(x - 1) (1 + x + . . . + x^{2013}) = 0$ $x^{2014} - 1 = 0$ $x^{2014} = 1 = e^{2 k π i}$ $\Rightarrow x = e^{\frac{2 k π i}{2014}} = e^{\frac{k π i}{1007}} = \cos \frac{k π}{1007} + i \sin \frac{k π}{1007}$ $w i t h k = 1, 2, . . ., 2013$ $i f x \in R :$ $x^{2014} = 1 \Rightarrow x = - 1$
	Commented by peter frank last updated on 05/Oct/22

	$thank you$
	Commented by Tawa11 last updated on 05/Oct/22

	$Great sir .$
	Answered by Strengthenchen last updated on 05/Oct/22

	$A s I t h o u g h t, t h e q u e s t i o n c o u l d b e s e e m a s a g e o m e t r i c p r o g r e s s i o n, d e a l i t l i k e t h i s :$ $\frac{1 (1 - x^{2014})}{1 - x} = 0$ $s o (1 - x^{2014}) = 0, e a s y t o k n o w x = \pm 1, 1 i s n o t f i t q u e s t i o n, - 1 i s f i n a l l y a n s w e r$
	Commented by mr W last updated on 06/Oct/22

	$a s f o r x \in C, 1 - x^{2014} = 0 h a s t o t a l l y$ $2014 r o o t s .$
	Answered by Rasheed.Sindhi last updated on 05/Oct/22

	$1 + x + x^{2} + x^{3} + . . . + x^{2012} + x^{2013} = 0$ $(1 + x) + x^{2} (1 + x) + . . . + x^{2012} (1 + x) = 0$ $(1 + x) (1 + x^{2} + x^{4} + . . . + x^{2012}) = 0$ $1 + x = 0 \Rightarrow x = - 1$
	Commented by peter frank last updated on 05/Oct/22

	$thank you$
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