# A more rigorous way to prove this? [duplicate]

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.

• You need $k=0$, not $n=0$. Apr 29, 2014 at 19:53
• Also, you need $x^k a^{(n-1)-k}$. See my answer below. Apr 29, 2014 at 20:10
• @MichaelHardy Thank you for pointing my mistake out. Pardon my quick typing. Apr 29, 2014 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. Apr 29, 2014 at 20:18

\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.