- #1

- 187

- 15

## Homework Statement

>Find the sum of the roots, real and non-real, of the equation [tex]x^{2001}+\left(\frac 12-x\right)^{2001}=0[/tex], given that there are no multiple roots.

While trying to solve the above problem (AIME 2001, Problem 3), I came across three solutions on [tex][AoPS](https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_3#See_also)[/tex]. But, I wonder if it could be solved as follows :

> Let the roots be [tex]P_1,P_2,...P_{2000}[/tex].

> The polynomial can be expressed as a product of factors as follows :

> [tex](1/2)(x-P_1)(x-P_2)....(x-P_{2000}) = 0[/tex].

> The above expression is the same as [tex]x^{2001}+\left(\frac 12-x\right)^{2001}=0[/tex].

>Thus, [tex]x^{2001}+\left(\frac 12-x\right)^{2001} = (1/2)(x-P_1)(x-P_2)....(x-P_{2000})[/tex]

> Here the coefficient of [tex]x^{1999}[/tex] on the RHS should represent ## \sum\limits_{i=1}^{2000}P_i*(-1/2)##.

> On the LHS the corresponding term would be the term with [tex]x^{1999}[/tex] and thus the coefficient of this term on the LHS should also be the required sum.

> On the LHS the coefficient of the [tex]x^{1999}[/tex] term is -##{2001}\choose{2})## into ##(1/2)^2## which represent the sum of the roots.

I have the following questions regarding the above :

1. Are there any inconsistencies in the reasoning?

2. The answers do not match, which seems to suggest so. (The answer through the methods on AoPS, using Vieta's is 500)

**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?**

Last edited: