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I am trying to understand why the following equation holds true:

$$\frac{x^{n}-1}{x+1} = x^{n-1} - x^{n-2} + \cdots - 1$$

where $n$ is even.

Using the long division, I was able to recover the first few terms on the right hand side, so I am convinced that it is true. But I would like to find an alternative argument. I think it might have something to do with power series, but I could not find the appropriate argument.

Josh
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2 Answers2

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Start with the geometric series formula. It's easy to find a proof of this online if you're not already familiar with it. $$1 + r + r^2 + r^3 + \cdots + r^{n-2} + r^{n-1} = \frac{1 - r^n}{1 - r}$$ Substitute $r = -x$, keeping in mind that since $n$ is even we have $r^n = x^n$. $$1 - x + x^2 - x^3 + \cdots + x^{n-2} - x^{n-1} = \frac{1-x^n}{1 + x}$$ Multiply both sides by $-1$ and you get your desired equality.

Trevor Kafka
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  • Please don't answer duplicates. Details on the reasons behind this request can be found here. – Matija Jan 15 '23 at 21:19
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    Given that it wasn't known to be a duplicate at the time of me posting this, I'm a bit confused why you're speaking to me like this as if I were knowledgably answering a duplicate. (Sidenote: it doesn't appear to be a duplicate as far as I can tell after reviewing your links.) – Trevor Kafka Jan 15 '23 at 21:23
  • No offense was intended, I'm convinced that you wanted to help and I only wanted to inform you about the guidelines, same as I was informed about them yesterday. – Matija Jan 15 '23 at 21:36
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I am confused with the inner-outer multlication proof. Let's take $n=4$: $$(x^3-x^2+x-1)\require{enclose}(x+1)=x^4+\enclose{horizontalstrike}{x^3-x^3}-\enclose{horizontalstrike} {x^2+x^2}+\enclose{horizontalstrike} {x-x}-1=x^4-1 $$ True! So is it true for any $n$?

Let's take $n=5:$ $$x^4-x^3+x^2-x+1...$$ I see... This is not in the given form. So, $n$ must be even. Then inner-outer multtplication proves the given identity, as in the proof of the well-known identity $\frac{x^n-1}{x-1}=x^{n-1}+x^{n-2}+...+x+1$.

Bob Dobbs
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