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Difference between revisions of "2008 iTest Problems/Problem 53"

(Solution to Problem 53 -- no worrying about most of the terms!)
 
(Fixed exponents)
 
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==Solution==
 
==Solution==
  
−
Because of [[Vieta's Formulas]], if we know the coefficient of the <math>x^2007</math> and <math>x^2006</math> term, we can find the sum of all the roots.  The coefficient of the <math>x^2007</math> term is easy to find -- it's <math>1</math>.  Using the [[Binomial Theorem]] in <math>(x-1)^2007</math>, the coefficient of the <math>x^2006</math> term is <math>-\tbinom{2007}{2006} + 2 = -2005</math>.  Thus, by Vieta's Formulas, the sum of all <math>2007</math> roots is <math>\tfrac{-(-2005)}{1} = \boxed{2005}</math>.
+
Because of [[Vieta's Formulas]], if we know the coefficient of the <math>x^{2007}</math> and <math>x^{2006}</math> term, we can find the sum of all the roots.  The coefficient of the <math>x^{2007}</math> term is easy to find -- it's <math>1</math>.  Using the [[Binomial Theorem]] in <math>(x-1)^{2007}</math>, the coefficient of the <math>x^{2006}</math> term is <math>-\tbinom{2007}{2006} + 2 = -2005</math>.  Thus, by Vieta's Formulas, the sum of all <math>2007</math> roots is <math>\tfrac{-(-2005)}{1} = \boxed{2005}</math>.
  
 
==See Also==
 
==See Also==

Latest revision as of 23:01, 12 July 2018

Problem

Find the sum of the $2007$ roots of $(x-1)^{2007}+2(x-2)^{2006}+3(x-3)^{2005}+\cdots+2006(x-2006)^2+2007(x-2007)$.

Solution

Because of Vieta's Formulas, if we know the coefficient of the $x^{2007}$ and $x^{2006}$ term, we can find the sum of all the roots. The coefficient of the $x^{2007}$ term is easy to find -- it's $1$. Using the Binomial Theorem in $(x-1)^{2007}$, the coefficient of the $x^{2006}$ term is $-\tbinom{2007}{2006} + 2 = -2005$. Thus, by Vieta's Formulas, the sum of all $2007$ roots is $\tfrac{-(-2005)}{1} = \boxed{2005}$.

See Also

2008 iTest (Problems)
Preceded by:
Problem 52 		Followed by:
Problem 54
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