Difference between revisions of "2007 Alabama ARML TST Problems/Problem 12"

Revision as of 12:54, 29 March 2009

If $w^{2007}=1$ and $w\neq 1$ , then evaluate

$\dfrac{1}{1+w}+\dfrac{1}{1+w^2}+\dfrac{1}{1+w^3}+\cdots +\dfrac{1}{1+w^{2007}}.$

Express your answer as a fraction in lowest terms.

For all $k \in \overline{1,2007}$ :

$x = w^k$ is a root of $f(x) = x^{2007} - 1$ .

$x = 1+w^k$ is a root of $g(x) = f(x-1) = (x-1)^{2007} - 1$ .

$x = \dfrac{1}{1+w^k}$ is a root of $h(x) = g\left(\dfrac{1}{x}\right) = \left(\dfrac{1}{x} - 1\right)^{2007} - 1$ .

$x = \dfrac{1}{1+w^k}$ is a root of $p(x) = x^{2007}h(x) = (1-x)^{2007} - x^{2007}$ , since $\dfrac{1}{1+w^k} \neq 0$ .

Since the degree of $p(x)$ is $2007$ , $p(x)$ has exactly $2007$ roots. So, the problem is asking for the sum of the roots of $p(x)$ .

Using the Binomial Theorem, $(1-x)^{2007} = -x^{2007} + 2007x^{2006} + \ldots + 1$ .

So, $p(x) = (1-x)^{2007} - x^{2007} = -2x^{2007} + 2007x^{2006} + \ldots + 1$ .

Therefore, by Vieta's Formulas, the sum of the roots of $p(x)$ is $-\dfrac{2007}{-2} = \boxed{\dfrac{2007}{2}}$ .

Revision as of 12:52, 29 March 2009 (view source) Xantos C. Guin (talk \| contribs) (added solution)		Revision as of 12:54, 29 March 2009 (view source) Xantos C. Guin (talk \| contribs) m Newer edit →
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−	Since the [[Degree_of_a_polynomial\|degree]] of <math>p(x)</math> is <math>2007</math>, <math>p(x)</math> has exactly <math>2007</math> roots, the problem is asking for the sum of the roots of <math>p(x)</math>.	+	Since the [[Degree_of_a_polynomial\|degree]] of <math>p(x)</math> is <math>2007</math>, <math>p(x)</math> has exactly <math>2007</math> roots. So, the problem is asking for the sum of the roots of <math>p(x)</math>.

	Using the [[Binomial Theorem]], <math>(1-x)^{2007} = -x^{2007} + 2007x^{2006} + \ldots + 1</math>.		Using the [[Binomial Theorem]], <math>(1-x)^{2007} = -x^{2007} + 2007x^{2006} + \ldots + 1</math>.