algebra precalculus - Algebraic Identity $a^{n}-b^{n} = (a-b) sumlimits_{k=0}^{n-1} a^{k}b^{n-1-k}$

Prove the following: $\displaystyle a^{n}-b^{n} = (a-b) \sum\limits_{k=0}^{n-1} a^{k}b^{n-1-k}$.

So one could use induction on $n$? Could one also use trichotomy or some type of combinatorial argument?

Answer

I have no idea what you mean by "use trichotomy," but here is the combinatorial argument. $a^n$ counts the number of words of length $n$ on the alphabet $\{ 1, 2, ... a \}$ and $b^n$ counts the number of words of length $n$ on the alphabet $\{ 1, 2, ... b \}$. Assume $a > b$. Then $a^n - b^n$ counts the number of words of length $n$ on the alphabet $\{ 1, 2, ... a \}$ such that at least one letter is greater than $b$.

Given such a word, suppose the last letter greater than $b$ occurs at position $k+1$. Then there are $a - b$ choices for this letter, $a^k$ choices for the letters before this letter, and $b^{n-k-1}$ choices for the letters after this letter. Thus there are $(a - b) a^k b^{n-k-1}$ such words, and summing over all $k$ gives

$$a^n - b^n = (a - b) \sum_{k=0}^{n-1} a^k b^{n-k-1}$$

as desired.