Using binomial coefficients to find sum of roots

  • #1
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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?
 
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Answers and Replies

  • #2
stevendaryl
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> Here the coefficient of [tex]x^{1999}[/tex] on the RHS should represent [tex]\sum\limits_{i=1}^{2000}P_i[/tex].
That's not correct. It should be ##- \frac{1}{2} \sum\limits_{i=1}^{2000}P_i##, shouldn't it?
 
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  • #3
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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!
 
  • #4
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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.
 

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