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I believe there's formula to write $1+x^{2n+1}$ in terms of $(1+x)(\cdots)$ just like how $1+x^3$ can be written as $(1+x)(x^2 -x +1)$. I am not able to find explanation of it anywhere over the internet.

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  • $\begingroup$ If you want to use binomial theorem (as original tag suggested), you can apply it on $1+x^{2n+1}=1+((x+1)-1)^{2n+1}$, giving eventually $1+x^{2n+1}=(1+x)\sum\binom{2n+1}{i}(-x-1)^{i-1}.$ But using geometric progression is definitely simpler. $\endgroup$
    – Sil
    Jul 14 '18 at 9:47
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This is more closely related to geometric series than binomial theorem.

First, recall the formula for a geometric series. In particular,

$$1 + x + x^2 + x^3 + \ldots + x^{n-1} = \frac{1 - x^n}{1 - x}.$$

Rearranging, this gives a factorisation result:

$$1 - x^n = (1 - x)(1 + x + x^2 + \ldots + x^{n-1}).$$

Now, substituting $-x$ in for $x$ (when $n$ is odd),

$$1 + x^n = (1 + x)(1 - x + x^2 - x^3 + \ldots + x^{n-1}).$$

Hope that helps!

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$$1+x^{2n+1}=(1+x)\sum_{j=0}^{2n}(-x)^j.$$This uses a geometric series, not the binomial theorem.

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