Will you really find roots?

The question is based on Binomial theorem ,

With help of it we will simplify the equation given to us.

$x^{2001}-(x-\frac{1}{2})^{2001}=0$

$x^{2001}-x^{2001}+\frac{2001}{2}x^{2000}-500250x^{1999} + ....=0$

$\frac{2001}{2}x^{2000}-500250x^{1999} + ....=0$

$x^{2000}-500x^{1999} + ....=0$

Hence, by Vieta's formula sum of roots is $\frac{-(-500)}{1}= \boxed{500}$

I did this a bit differently, using symmetry considerations:

Let $x=\frac{1}{4}+y$ . The polynomial becomes $(\frac{1}{4}+y)^{2001}+(\frac{1}{4}-y)^{2001}$ , and it is fairly easy to see that its roots, whether real or complex, must be symmetrical around $\frac{1}{4}$ . Each symmetrical pair of roots adds up to $\frac{1}{2}$ , and there are $2,000$ roots in the complex plane since the polynomial has degree $2,000$ (note that the degree- $2001$ terms cancel). So the answer must be $\frac{1}{4}*2,000=\frac{1}{2}*1,000=\boxed{500}$ .

More generally, for any real number $m$ and for any odd integer $n$ , the sum of the complex roots of $x^{n}+(m-x)^{n}$ ought to be $\frac{(n-1)*m}{2}$ , and for any even integer $n$ it ought to be $\frac{n*m}{2}$ .