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I'm trying to solve this question:

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$$S_n = 1+x+x^2+x^3...+x^{2021}$$ $$xS_n = x+x^2+x^3+x^4...+x^{2022}$$ Subtracting the bottom equation from the top.. $$S_n(1-x)=1-x^{2022}$$ $$S_n=\dfrac{1-x^{2022}}{1-x}$$

Giving that $S_n = 0$, I ended up with this equation: $\dfrac{1-x^{2022}}{1-x}=0$

When graphed, there is only one real solution, which is $-1$, as long as $x$ is raised to an even power. How can I solve that equation to get $-1$? That's my question.

Angelo
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mjo
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    Do you know the method to find the $n$th roots of a complex number? Apply it to $1$, get its $n$th roots. – GEdgar Sep 17 '22 at 01:05
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    Multiply both sides by $1 - x$ to get $1 - x^n = 0$, or $x^n = 1$. Take the $n$th root of both sides to get your solution (taking into account $\pm$ when $n$ is even). Don't forget, your method excludes $x = 1$ from being a solution, so you should deal with that case via a separate argument. – Theo Bendit Sep 17 '22 at 01:11
  • @TheoBendit Okay, the last point made it clear for me. I had thought of it, but when I got +1 as a solution, I assumed something was wrong. Thanks! – mjo Sep 17 '22 at 01:16

2 Answers2

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Hint Factor the sum as $$S = (1 + x)(1 + x^2 + \cdots + x^{2020}) .$$

Travis Willse
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The solutions are the $n-1$ $n$-th roots of unity other than $1$. Those are $e^{2\pi ik/n},\,k=1,\dots n-1$. For $n$ even, we have $e^{\pi i}=-1$ as the only real one.

That's other than $k=n/2$, they all have nonzero imaginary part $\sin(2\pi k/n)$.

(The $n$-th roots of unity are evenly spaced on the unit circle, if you know anything about the circle group.)

calc ll
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