2001 AIME I Problems/Problem 3

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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
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All AIME Problems and Solutions
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