Using binomial coefficients to find sum of roots

[JC2000]

Homework Statement

>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.

While trying to solve the above problem (AIME 2001, Problem 3), I came across three solutions on $$[AoPS](<a href="https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_3#See_also)$$" target="_blank" class="link link--external" rel="nofollow ugc noopener">https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_3#See_also)[/tex]</a>. But, I wonder if it could be solved as follows :<br /> <br /> > Let the roots be $$P_1,P_2,...P_{2000}$$.<br /> > The polynomial can be expressed as a product of factors as follows :<br /> > $$(1/2)(x-P_1)(x-P_2)...(x-P_{2000}) = 0$$.<br /> > The above expression is the same as $$x^{2001}+\left(\frac 12-x\right)^{2001}=0$$.<br /> <br /> >Thus, $$x^{2001}+\left(\frac 12-x\right)^{2001} = (1/2)(x-P_1)(x-P_2)...(x-P_{2000})$$<br /> <br /> > Here the coefficient of $$x^{1999}$$ on the RHS should represent $\sum\limits_{i=1}^{2000}P_i*(-1/2)$.<br /> <br /> > On the LHS the corresponding term would be the term with $$x^{1999}$$ and thus the coefficient of this term on the LHS should also be the required sum.<br /> <br /> > On the LHS the coefficient of the $$x^{1999}$$ term is -${2001}\choose{2})$ into $(1/2)^2$ which represent the sum of the roots.<br /> <br /> I have the following questions regarding the above :<br /> <br /> 1. Are there any inconsistencies in the reasoning?<br /> 2. The answers do not match, which seems to suggest so. (The answer through the methods on AoPS, using Vieta's is 500)<br /> <b>3. Is there a way of arriving at the answer without using Vieta's formula and by expressing the polynomial as a product of factors and then using binomial coefficients as attempted above?</b>[/tex]

 

[stevendaryl]

> Here the coefficient of $$x^{1999}$$ on the RHS should represent $$\sum\limits_{i=1}^{2000}P_i$$.

That's not correct. It should be $- \frac{1}{2} \sum\limits_{i=1}^{2000}P_i$, shouldn't it?

 

[JC2000]

That's not correct. It should be $- \frac{1}{2} \sum\limits_{i=1}^{2000}P_i$, shouldn't it?

Oh yes! After considering that, $\sum\limits_{i=1}^{2000}P_i$ should be 1,000,500! Thank you!

 

[JC2000]

Thanks for your time! The error in this solution was explained to me elsewhere. It seems that the leading coefficient for the factorization on RHS should actually be 2001/2 instead of 1/2. Making that changes gives the correct answer.