Why does $ \frac{b^n-a^n}{b-a}=\sum_{k=1}^nb^{n-k}a^{k-1}$?

Question

Trying to work through the answer in this question:

The inequality $b^n - a^n < (b - a)nb^{n-1}$

Answer 1

There are a couple of ways to do this:

1) Factorizing $b^n-a^n$: For example if $n=2$- then $b^2-a^2=(b-a)(b+a)$."Long divison" gives the answer for general $n$.

2) Multiply RHS with $(b-a)$ to get $b^n-a^n$. There would be a lot of cancellations, and you will be left with $b^n-a^n$.

In some sense 1) and 2) are the "same" methods- though the flavor is a bit different.

Answer 2

First consider a sample of the equation to be shown. Consider \begin{align} a^{2} + ab + b^{2} &= \frac{(b-a)(b^{2} + ab + a^{2})}{b-a} \\ &= \frac{1}{b-a} \, [ b^{3} -a b^{2} + a b^{2} - a^{2} b + a^{2} b - a^{3} = \frac{b^{3}- a^{3}}{b-a}. \end{align} Now apply the same logic to the form \begin{align} b^{n-1} + a b^{n-2} + \cdots + a^{n-2} b + a^{n-1} \end{align} for which the result is \begin{align} \frac{b^{n} - a^{n}}{b-a} = \sum_{k=0}^{n-1} b^{n-k-1} \, a^{k}. \end{align}