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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, in a polynomial of the form a_nx^n + a_{n-1}x^{n-1} + \cdots + a_0 = 0, then the sum of the roots is \frac{-a_{n-1}}{a_n}.

From the Binomial Theorem, the first term of \left(\frac 12-x\right)^{2001} is -x^{2001}, but x^{2001}+-x^{2001}=0, so the term with the largest degree is x^{2000}. So we need the coefficient of that term, as well as the coefficient of x^{1999}.

\begin{align*}\binom{2001}{1} \cdot (-x)^{2000} \cdot \left(\frac{1}{2}\right)^1&=\frac{2001x^{2000}}{2}\\\binom{2001}{2}...

Applying Vieta's formulas, we find that the sum of the roots is -\frac{-2001 \cdot 250}{\frac{2001}{2}}=250 \cdot 2=\boxed{500}.

See also

2001 AIME I (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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