I need to know if the binomial theorem can somehow be applied to: $$\frac{a^n - b^n}{a- b}$$ I've done a bit of research but I still don't know where to begin with this one so I can't offer any pointers. Any suggestions?

Edit: thanks for all replies! The question was provoked by this reddit post, which was related to calculus. The most relevant SE post is this one here.

  • 1
    $\begingroup$ It looks similar to the sum of geometric series. $\endgroup$
    – Chinny84
    Dec 17, 2015 at 12:51

3 Answers 3


$$\frac{a^n-b^n}{a-b} = \frac{(a-b)\sum_{k=1}^{n} a^{n-k}b^{k-1}}{a-b} = \sum_{k=1}^{n} a^{n-k}b^{k-1}$$

Using the familiar "difference of two powers" factorisation.

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    $\begingroup$ I would conclude "No, you cannot use the binomial theorem, because what you get is a sum similar to that appearing in the binomial formula, but without the binomial coefficients". $\endgroup$ Dec 17, 2015 at 12:55

Perhaps an overkill (compared to other answers), but since the geometric sum was mentioned in the comments: $$\frac{a^n-b^n}{a-b}=\frac{a^n\left(1-\left(\frac{b}{a}\right)^n\right)}{a\left(1-\frac{b}{a}\right)}=a^{n-1}\cdot\frac{1-\left(\frac{b}{a}\right)^n}{1-\frac{b}{a}}=a^{n-1}\sum_{k=0}^{n-1}\left(\frac ba\right)^k$$


Let c = a - b.

Hence, a = b + c.

Hence, $$ \frac{a^n - b^n}{a- b} $$

becomes $$ \frac{(b+c)^n - b^n}{c} $$

The numerator can be expanded using binomial theorem. The first term of the expansion will be $b^n$ which will get cancelled out. All other terms have c as a factor which will cancel out with the denominator.

Hope that helps.

  • $\begingroup$ But why do you want to apply the binomial theorem to it in the first place? replacing c with a-b in the end result will anyway lead to elimination of binomial coefficients as pointed out by the other answers. $\endgroup$ Dec 17, 2015 at 13:05
  • $\begingroup$ The final expression would be a sum of products of powers of $b$ and $c$, i.e. $b$ and $a-b$, and this is far more complex that the initial expression. Why to do so? $\endgroup$ Dec 17, 2015 at 13:40
  • $\begingroup$ I already pointed that out in my own comment to my answer. The reason I showed it was to show how binomial theorem could be used here as was specifically asked for. And we need not worry about the complexity of the resulting expansion as we already know how the end result is going to look like. $\endgroup$ Dec 17, 2015 at 14:37
  • $\begingroup$ @Deepak Gupta I accepted your answer because it answers the question most directly. However, as you seem to say yourself, it may not be the best or simplest solution. The edit in the post contains information on the background to the problem. $\endgroup$
    – user117644
    Dec 17, 2015 at 22:04

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