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How can we prove that the following expression is a polynomial? $$ \frac{1-x^{2^{n-1}}}{1-x} $$ I've asked this question just for learning the ways different from using

$1-x^n=(1-x)(1+x+x^2+···+x^{n-1})$.

frosh
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3 Answers3

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Hint: for any $k\in\Bbb N$ we have $1-x^k=(1-x)(1+x+x^2+x^3+...+x^{k-1})$.

Wojowu
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    Even easier, $1-1^{2^{n-1}}=0$, so $1-x^{2^{n-1}}=(x-1)P(x)$ for some polynomial $P$. We don't need a constructive solution ;) – Cody Johnson May 20 '15 at 20:18
  • @feynman Not quite - in your case $k=2^{n-1}$, so last exponent would be $2^{n-1}-1$. – Wojowu May 20 '15 at 20:21
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    @Cody Johnson Your comment is a great perspective on this question. You should turn that into an answer! – Hrodelbert May 20 '15 at 20:22
  • Did I do it correctly? Sorry, I'm new to this website – Cody Johnson May 20 '15 at 20:29
  • @CodyJohnson Yes, you did it correctly: you are free to comment on answers. But if your comment entails a significantly different and interesting view on the problem at hand, sometimes someone comes along to ask you to write an answer about it (by using the box below). Right now, this is no longer possible for this question, since it is been labeled as duplicate. Maybe next time;) – Hrodelbert May 21 '15 at 07:53
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Even easier, $1-1^{2^{n-1}}=0$, so $1-x^{2^{n-1}}=(x-1)P(x)$ for some polynomial $P$. We don't need a constructive solution ;)

Cody Johnson
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We have

$$\forall m\in \mathbb{N}, 1-x^m=(1-x)(1+x+\cdots+x^{m-1})$$

Just take $m=2^{n-1}$ and you're done

marwalix
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