Prove $\frac{a^n−b^n}{a−b}=\sum _{k=0} ^{n−1} a^kb^{n−1−k}$ by Induction

Question

Let $a\ne b\in\Bbb R$. Show that for all $n\in\mathbb{N}$:

$$\frac{a^n−b^n}{a−b}=\sum _{k=0} ^{n−1} a^kb^{n−1−k}$$

I know using induction proof to prove it, but I can do base case but I am having trouble on proving if $n$ is true then $n+1$ is true. Please give me some hints on this. I tried to use $n+1 - n$ but I can't let the left side equal to the right side. My knowledge I’d only induction proof and a little bit of binomial theory.

Answer 1

Hint: if $$f_n=\frac{a^n-b^n}{a-b},$$ you need a way to go from $f_n$ to $f_{n+1}$. Now $$a^{n+1}-b^{n+1}=a\,\left(a^n-b^n\right)+(a-b)\,b^n,$$i.e. $$f_{n+1}=a\,f_n+b^n.$$

Answer 2

I think you can prove this by inspection, just multiplying

\begin{align*} a^n-b^n = (a-b)(b^{n-1}+b^{n-2}a+b^{n-3}a^2+\dots+a^{n-1}) \end{align*}

This appears in a book from Elias Stein, Complex Analysis.