The remainder when $1 + x + x^{2} + x^{3} + . . . + x^{2006}$ is divided by $(x - 1)$ is :

2005

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2006

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2007

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2008

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Solution

The correct option is D $2007$
The remainder $= f (1)$ by remainder theorem.
$∴$ Remainder $= 1 + 1^{2} + 1^{3} + 1^{4} + . . . . . . + 1^{2006}$
$∴$ Remainder $= 1 + 1 + 1 + 1.... + 1 (2007$ times) $= 2007$

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