I would like to prove the following statement $$x^n-a^n=(x-a)\sum^{n-1}_{k=0}x^ka^{n-k-1},\qquad\forall n\in\Bbb N_0$$ I can easily prove it by induction using polynomial long division or series expansion however I am unsure whether or not these are "rigorous enough". I have thought about this for a while but is there a way to prove this statement in a more rigorous way? I think that the use of ellipses and "do this for the remaining $n$" gives me this uncertainty. Therefore I pose my question as follows:

Is there a rigorous way to prove this statement, or is the use of ellipses rigorous enough?

Thank you.

marked as duplicate by lhf, ml0105, egreg, Git Gud, colormegone Apr 29 '14 at 21:29

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  • 1
    You need $k=0$, not $n=0$. – Michael Hardy Apr 29 '14 at 19:53
  • Also, you need $x^k a^{(n-1)-k}$. See my answer below. – Michael Hardy Apr 29 '14 at 20:10
  • @MichaelHardy Thank you for pointing my mistake out. Pardon my quick typing. – Ali Caglayan Apr 29 '14 at 20:11
  • I don't think this is a duplicate. It may be similar to the possible duplicate but my question is specifically asking about rigor as well, which the other question does not talk about. – Ali Caglayan Apr 29 '14 at 20:18
up vote 1 down vote accepted

\begin{align} & (x-a)\sum^{n-1}_{k=0}x^ka^{n-1-k} \\[10pt] = & x \sum^{n-1}_{k=0}x^ka^{n-1-k} - a \sum^{n-1}_{k=0}x^ka^{n-1-k} \\[10pt] = & \sum_{k=0}^{n-1} x^{k+1}a^{n-1-k} - \sum_{k=0}^{n-1} x^k a^{n-1-k+1} \\[10pt] = & \sum_{\ell=1}^n x^\ell a^{n-\ell} - \sum_{k=0}^{n-1} x^k a^{n-k}. \end{align} Here we've let $\ell=k+1$. But after that it does not matter whether the index that runs from $1$ to $n$ is called $\ell$ or $k$, so we have \begin{align} & \sum_{k=1}^n x^k a^{n-k} - \sum_{k=0}^{n-1} x^k a^{n-k} \\[10pt] = & \left(\left( \sum_{k=1}^{n-1} x^k a^{n-k}\right)+ x^n a^0 \right) - \left( x^0 a^{n-0} + \sum_{k=1}^{n-1} x^k a^{n-k} \right) \\[10pt] = & x^n - a^n \end{align}

Develop and change the index: (almost all the terms are canceled except for $k=0$ and $k=n$):

$$(x-a)\sum^{n-1}_{k=0}x^ka^{n-1-k}=\sum^{n-1}_{k=0}x^{k+1}a^{n-1-k}-\sum^{n-1}_{k=0}x^ka^{n-k}=\sum^{n}_{k=1}x^{k}a^{n-k}-\sum^{n-1}_{k=0}x^ka^{n-k}\\=x^n-a^n$$ but I would say that all the proofs you mentioned are also rigorous.

Yet another way: \begin{eqnarray*} (x-a)\sum_{k=0}^{n-1}x^{k}a^{n-k-1} &=&(x-a)a^{n-1}\sum_{k=0}^{n-1}(\frac{x}{% a})^{k} \\ &=&(x-a)a^{n-1}\frac{1-(\frac{x}{a})^{n}}{1-\frac{x}{a}}=(x-a)\frac{% a^{n}-x^{n}}{a-x}=x^{n}-a^{n} \end{eqnarray*}

Using a series expansion is definitely rigorous enough. It's the most straightforward way to demonstrate the equality, using only basic algebra.

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