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Pythagoras Triplets

If a satisfie...

Question

If a satisfies the equation $a^{2017} - 2 a + 1 = 0$ and $S = 1 + a + a^{2} + . . . + a^{2016}$ , then possible value(s) of S is

2016

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2018

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2017

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Solution

The correct option is D 2017
solution of $a^{2017} - 2 a + 1 = 0$ is $a = 1$
putting $a = 1$
$S = 1 + 1 + 1^{2} + . . . . . . . . . . . . . . . . . . . {.1}^{2016} = 1 + (2016) = 2017$
Therefore Answer is $C$

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