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)[/tex]" 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 /> &gt; Let the roots be P_1,P_2,...P_{2000}.<br /> &gt; The polynomial can be expressed as a product of factors as follows :<br /> &gt; (1/2)(x-P_1)(x-P_2)...(x-P_{2000}) = 0.<br /> &gt; The above expression is the same as x^{2001}+\left(\frac 12-x\right)^{2001}=0.<br /> <br /> &gt;Thus, x^{2001}+\left(\frac 12-x\right)^{2001} = (1/2)(x-P_1)(x-P_2)...(x-P_{2000})<br /> <br /> &gt; Here the coefficient of x^{1999} on the RHS should represent $\sum\limits_{i=1}^{2000}P_i*(-1/2)$.<br /> <br /> &gt; 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 /> &gt; 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&#039;s is 500)<br /> <b>3. Is there a way of arriving at the answer without using Vieta&#039;s formula and by expressing the polynomial as a product of factors and then using binomial coefficients as attempted above?</b>

 

[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.