Difference between revisions of "2001 AIME I Problems/Problem 3"

m
Line 3: Line 3:
  
 
== Solution ==
 
== Solution ==
{{solution}}
+
From [[Vieta's formulas]], we just need to find the first two terms.
  
 +
From the [[Binomial Theorem]], the first term of <math>left(\frac 12-x\right)^{2001}</math> is <math>-x^{2001}</math>, but <math>x^{2001}+-x^{2001}=0</math>, so the first term has <math>x^{2000}</math> in it, not <math>x^{2001}</math>. So we find that term, and the term with <math>x^{1999}</math>.
 +
 +
<math>x^{2000}*\binom{2001}{1}*\frac{1}{2}=\frac{2001x^{2000}}{2}</math>
 +
 +
<math>-x^{1999}*\binom{2001}{2}*\frac{1}{4}=\frac{-x^{1999}*2001*2000}{8}=-x^{1999}2001*250</math>
 +
 +
Applying Vieta's Formulas, we get that the sum of the roots is
 +
 +
<math>-\frac{-2001*250}{\frac{2001}{2}}=250*2=\boxed{500}</math>
 
== See also ==
 
== See also ==
 
{{AIME box|year=2001|n=I|num-b=2|num-a=4}}
 
{{AIME box|year=2001|n=I|num-b=2|num-a=4}}

Revision as of 13:43, 1 December 2007

Problem

Find the sum of the roots, real and non-real, of the equation $x^{2001}+\left(\frac 12-x\right)^{2001}=0$, given that there are no multiple roots.

Solution

From Vieta's formulas, we just need to find the first two terms.

From the Binomial Theorem, the first term of $left(\frac 12-x\right)^{2001}$ (Error compiling LaTeX. ! Extra \right.) is $-x^{2001}$, but $x^{2001}+-x^{2001}=0$, so the first term has $x^{2000}$ in it, not $x^{2001}$. So we find that term, and the term with $x^{1999}$.

$x^{2000}*\binom{2001}{1}*\frac{1}{2}=\frac{2001x^{2000}}{2}$

$-x^{1999}*\binom{2001}{2}*\frac{1}{4}=\frac{-x^{1999}*2001*2000}{8}=-x^{1999}2001*250$

Applying Vieta's Formulas, we get that the sum of the roots is

$-\frac{-2001*250}{\frac{2001}{2}}=250*2=\boxed{500}$

See also

2001 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
Invalid username
Login to AoPS