Difference between revisions of "2001 AIME I Problems/Problem 3"
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== 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>. | ||
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+ | <math>x^{2000}*\binom{2001}{1}*\frac{1}{2}=\frac{2001x^{2000}}{2}</math> | ||
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+ | <math>-x^{1999}*\binom{2001}{2}*\frac{1}{4}=\frac{-x^{1999}*2001*2000}{8}=-x^{1999}2001*250</math> | ||
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+ | Applying Vieta's Formulas, we get that the sum of the roots is | ||
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+ | <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 , 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 , but , so the first term has in it, not . So we find that term, and the term with .
Applying Vieta's Formulas, we get that the sum of the roots is
See also
2001 AIME I (Problems • Answer Key • Resources) | ||
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 |