# What is the remainder when 1 + x + x^2 + x^3 + ... + x^2006 is divided by (x-1) ?

Dec 4, 2016

$2007$

#### Explanation:

If $f \left(x\right)$ is a polynomial, then the remainder when dividing $f \left(x\right)$ by $\left(x - a\right)$ is $f \left(a\right)$

In our example:

$f \left(x\right) = 1 + x + {x}^{2} + {x}^{3} + \ldots + {x}^{2006}$

So the remainder when dividing by $\left(x - 1\right)$ is:

$f \left(1\right) = 1 + 1 + 1 + 1 + \ldots + 1 = 2007$

Dec 10, 2016

$2007$

#### Explanation:

$1 + x + {x}^{2} + \cdots + {x}^{2006} = \frac{{x}^{2007} - 1}{x - 1} = q \left(x\right) \left(x - 1\right) + C$

or

${x}^{2007} - 1 = q \left(x\right) {\left(x - 1\right)}^{2} + C \left(x - 1\right)$ now deriving both sides

$2007 {x}^{2006} = 2 q \left(x\right) \left(x - 1\right) + q ' \left(x\right) {\left(x - 1\right)}^{2} + C$

Making now $x = 1$ we have

$2007 = C$

so the answer is $2007$