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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?

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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.

Someone should mention the “polynomial multiplication” or “telescoping” proof, which may be viewed as a variant of the “geometric series” method.

$$\begin{align*} (a-b)\sum_{k=0}^{n-1} a^k b^{n-1-k} &= \sum_{k=0}^{n-1} a^{k+1} b^{n-1-k} – \sum_{k=0}^{n-1} a^k b^{n-k} \\ &= \sum_{k=1}^n a^k b^{n-k} – \sum_{k=0}^{n-1} a^k b^{n-k} = a^n – b^n. \end{align*}$$

EDIT

**Proof by induction**

$n=1$ is valid.

Supose valid by n, then

$$a^{n+1}-b^{n+1}=a(a^{n})+b(b^{n})$$, using the hipotesis :

$$a(a^{n})+b(b^{n})=a\left[b^{n}+(b-a)\displaystyle\sum\limits_{k=0}^{n-1} a^{k}b^{n-1-k}\right] + b\left[a^{n}-(b-a)\displaystyle\sum\limits_{k=0}^{n-1} a^{k}b^{n-1-k}\right]=$$

$$\left[ab^{n}+(b-a)a\displaystyle\sum\limits_{k=0}^{n-1} a^{k}b^{n-1-k}\right] + \left[a^{n}b-(b-a)b\displaystyle\sum\limits_{k=0}^{n-1} a^{k}b^{n-1-k}\right]=$$

$$\left[ab^{n}+(b-a)\displaystyle\sum\limits_{k=0}^{n-1} a^{k+1}b^{n-1-k}\right] + \left[a^{n}b-(b-a)\displaystyle\sum\limits_{k=0}^{n-1} a^{k}b^{n-k}\right]=$$

$$\left[ab^{n}+(b-a)\displaystyle\sum\limits_{k=0}^{n-1} a^{k+1}b^{n-1-k}+(b-a)b^{n}-(b-a)b^{n}\right] +$$

$$ \left[a^{n}b-(b-a)\displaystyle\sum\limits_{k=0}^{n-1} a^{k}b^{n-k}-(b-a)a^{n}+(b-a)a^{n}\right]=$$

$$\left[(b-a)[\displaystyle\sum\limits_{k=0}^{n-1} a^{k+1}b^{n-1-k}+b^{n}]+b^{n+1}\right] +\left[(b-a)[\displaystyle\sum\limits_{k=0}^{n-1} a^{k}b^{n-k}+a^{n}]-a^{n+1}\right] +$$

$$\left[(b-a)\displaystyle\sum\limits_{k=0}^{n} a^{k}b^{n-k}+b^{n+1}\right] +\left[(b-a)\displaystyle\sum\limits_{k=0}^{n} a^{k}b^{n-k}-a^{n+1}\right] =$$

$$-a^{n+1}+b^{n+1}+2\left[(b-a)\displaystyle\sum\limits_{k=0}^{n} a^{k}b^{n-k}\right] =$$

Thus:

$$a^{n+1}-b^{n+1}=-a^{n+1}+b^{n+1}+2\left[(b-a)\displaystyle\sum\limits_{k=0}^{n} a^{k}b^{n-k}\right] $$, then

$$2(a^{n+1}-b^{n+1})=2\left[(b-a)\displaystyle\sum\limits_{k=0}^{n} a^{k}b^{n-k}\right]$$

thus:

$$a^{n+1}-b^{n+1}=\left[(b-a)\displaystyle\sum\limits_{k=0}^{n} a^{k}b^{n-k}\right]$$

So $n+1$ is valid.

Complete the proof

You can apply Ruffini’s rule. Here is a copy from my Algebra text book (Compêndio de Álgebra, VI, by Sebastião e Silva and Silva Paulo) where the following formula is obtained:

$x^n-a^n\equiv (x-a)(x^{n-1}+ax^{n-2}+a^2x^{n-3}+\cdots +a^{n-2}x+a^{n-1}).$

Translation: The Ruffini’s rule can be used to find the quotient of $x^n-a^n$ by $x-a$:

(Figure)

Thus, if $n$ is a natural number, we have

$x^n-a^n\equiv (x-a)(x^{n-1}+ax^{n-2}+a^2x^{n-3}+\cdots +a^{n-2}x+a^{n-1})$

Yes, you could use induction on n. I don’t see an easy trichotomy or combinatorial argument.

Can we build a combinatorial argument along these lines?

Say if we have say $n$ students to be allotted in $a$ rooms with $b$ of the $a$ room being non air-conditioned. (Assume the students are distinguishable so we can order the student as $1,2,3,…,n$)

The total number of ways is $a^n$.

Suppose all students are allotted to the $b$ rooms, the number of ways is $b^n$.

If the first $n-1$ students are allotted to the $b$ rooms, and the final dude in some other room, the number of possible ways is $b^{n-1} \times (a-b)$.

Now if the first $n-2$ students are allotted to the $b$ rooms, and the remaining two students are now left. If the $(n-1)^{th}$ student chooses from the $b$ rooms then we are back to the earlier case. So the $(n-1)^{th}$ student needs to choose from the remaining $(a-b)$ rooms. Now the $n^{th}$ student can choose from any of the $a$ rooms. The number of possible ways is $b^{n-2} \times (a-b) \times a$.

In general, if the first $n-k$ students are allotted to the $b$ rooms, and the remaining $k$ students are now left. If the $(n-k+1)^{th}$ student chooses from the $b$ rooms then we are back to the previous case. So the $(n-k+1)^{th}$ student needs to choose from the remaining $(a-b)$ rooms. Now the students from $(n-k+2)$ to $n$ can choose from any of the $a$ rooms. The number of possible ways is $b^{n-k} \times (a-b) \times a^{k-1}$.

So, the total number of ways is $b^n + \displaystyle \sum_{k=1}^n b^{n-k} \times (a-b) \times a^{k-1}$.

Both the counting must add up and hence we get $a^n = b^n + \displaystyle \sum_{k=1}^n b^{n-k} \times (a-b) \times a^{k-1}$.

You could use geometric series to conclude the result as well.

The right hand side is $(a-b) b^{n-1} \displaystyle \sum_{k=0}^{n-1} (\frac{a}{b})^k = (a-b) b^{n-1} \frac{((\frac{a}{b})^n – 1)}{(\frac{a}{b})-1} = (a-b) \frac{a^n – b^n}{a-b} = a^n – b^n$

If you know the sum of a geometric sequence, then set $x=a/b$ and conclude

$

x^n-1 = (x-1)(1+x+x^2+\cdots+x^{n-1})

$. Now multiply by $b^n$. (If $b=0$, then the identity is obvious.)

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