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Find the sum of the roots, real and non-real, of the equation $$ x^{2001} + \left( \frac{1}{2} - x \right)^{2001} = 0 $$ given that there are no multiple roots.

I am in a weird situation here.

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

$$x = -\frac{1}{2} + x$$

Then I get $0 = -\frac{1}{2}$ which I know is not true.

So all roots have to be complex roots. By vieta's formulas:

$$\sum _{k=1}^{n} r_k = -\frac{a_{n-1}}{a_n}$$

But that cant be of help here. The leading coefficient is $Cx^{2000}$.

Hints please, no complete answers, thank you!

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    $\begingroup$ well can't you use Newton's binomial theorem to compute the leading coefficients? $\endgroup$ – Surb Jun 29 '15 at 18:04
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    $\begingroup$ Ah jesus. $$(1/2 - x)^{2001} = \sum_{k=0}^{2001} \binom{2001}{k} (1/2)^k (-x)^{2001 - k}$$ the $k's$ I am after are: $k = \{1, 2\}$. Hence, $a_n = 2001(1/2)$ and $a_{n-1} = 2001000 (1/4) = 500250$. $\sum r_k = \frac{1000500}{2001} = 500$$? $\endgroup$ – Amad27 Jun 29 '15 at 19:17
  • $\begingroup$ You're on the good track :). Pay attention to the fact that $a_n = 2001\cdot (1/2)+\color{red}{1}$ (since you are dealing with $\color{red}{x^{2001}}+(1/2-x)^{2001}$) Moreover note that your sum starts at $0$ (and so $(1/2)^0 = 1$) and don't forget the $\color{blue}{-}$ in $(\color{blue}{-}x)^{2001-k}=\color{blue}{(-1)^{2001-k}}x^{2001-k}$ $\endgroup$ – Surb Jun 29 '15 at 20:40
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$x^{2001} + (\frac{1}{2} - x)^{2001} = 0 \implies x^{2001} + \sum_0 ^{2001} \binom{2001}{k} (\frac{1}{2})^{2001 - k} (-x)^k = 0$... What does Vieta's formula say at this point?

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HINT: $(z^a)^{\frac{1}{a}}\neq z$ In general. There are exactly $n$ $n^{th}$ roots of a number if we allow complex numbers.

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Try using a difference of two powers formula on $x^{2001} - (x - 1/2)^{2001}$, then using Vieta's formulae.

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