The original question is the following:

Because of the complete symmetry between $x$ and $\dfrac{1}{2}-x$, the roots $r$ and $\dfrac{1}{2}-r$ should have the same multiplicity. If you want to be more rigorous, then Vieta's Formulas can be applied to calculate the sum of the roots.

Consider the binomial expansion of the second part. It contains $-x^{2001}$, so the term $x^{2001}$ is canceled, and the highest degree is $2000$.  Then it is a polynomial of degree $2000$, and there are $2000$ roots. Now the coefficient of the term $x^{2000}$ is $\displaystyle\binom{2001}{2000}\frac{1}{2}=\frac{2001}{2}$, and the coefficient of the term $x^{1999}$ is $\displaystyle -\binom{2001}{1999}\frac{1}{2^2}=-\frac{2001\cdot 2000}{8}$. Therefore, by Vieta's formulas, the sum of the roots is