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Is it possible to come up with a combinatorial argument which proves the following identity?

$$a^n - b^n = (a - b)(a^{n – 1} + a^{n – 2}b + \dots + ab^{n – 2} + b^{n – 1})$$

My idea was this:

Consider the number of ways $N$ to distribute $n$ numbered balls into $a$ numbered boxes such that there must be at least $1$ ball among the first $a - b$ boxes. The number of ways to do this is take the total number of ways without restriction, and subtract off the number of ways that all the balls are distributed only among the last $b$ boxes. That is,

$$N = a^n - b^n$$

Alternatively, we can first pick at least $1$ ball(s) and distribute them among the first $a - b$ boxes, and then distribute the remaining balls among the last $b$ boxes. That is,

$$N = \binom{n}{1}(a-b)b^{n-1} + \binom{n}{2}(a-b)^2b^{n-2} + \dots + \binom{n}{n}(a-b)^nb^0$$

However, this does not resemble the required $RHS$ of the identity.

Yiyuan Lee
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    http://math.stackexchange.com/questions/900739/how-many-different-proofs-are-there-that-an-bn-a-b-sum-i-0n-1-ai-b/900782#900782 – SuperAbound Aug 17 '14 at 10:55
  • Thank you @Superabound, but the answer linked does not provide any combinatorial argument... – Yiyuan Lee Aug 17 '14 at 10:56

2 Answers2

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Let ball $k$ be the highest numbered ball that gets placed in one of the first $a-b$ boxes. Then all balls numbered $k+1$ through $n$ get placed in the last $b$ boxes, whereas balls numbered $1$ through $k-1$ may be placed in any of the $a$ boxes.

Will Orrick
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  • What if the even balls go into the first $a - b$ boxes while the odd balls into the last $b$ boxes? – Yiyuan Lee Aug 17 '14 at 11:20
  • I'm uncertain how your question relates to my answer. The last even numbered ball is ball $n$ if $n$ is even and ball $n−1$ if $n$ is odd. The configurations you describe are among the $(a−b)a^{n−1}$ possible configurations where $n$ is even and ball $n$ goes into one of the first $a−b$ boxes. They're among the $(a−b)a^{n−2}b$ possible configurations where $n$ is odd and ball $n−1$ goes into one of the first $a−b$ boxes while ball $n$ goes into one of the last $b$ boxes. – Will Orrick Aug 18 '14 at 15:00
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Using the binomial formula:

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

And $$\sum_{k=0}^{n-1}{n\choose k}(a-b)^{n-k-1}b^k = \sum_{k=0}^{n-1} \sum_{j=0}^{n-k-1}{n\choose k}{n-k-1\choose j}(-1)^ja^{n-k-1-j}b^{j+k} = \sum_{i=0}^{n-1}(\sum_{j+k=i} {n\choose k}{n-k-1\choose j}(-1)^j)a^{n-1-i}b^i$$

We need to prove $\sum_{j+k=i} {n\choose k}{n-k-1\choose j}(-1)^j = 1$ i.e. $ \sum_{j=0}^i{i\choose j}\frac{(-1)^j}{n-(i-j)} = \dfrac{1}{{n \choose i}(n-i)}$

This can be proven by partial fraction decomposition combined with the fact $\sum_{j=0}^n(-1)^{n-j}{n\choose j} = 0$.

The partial decomposition says \begin{align} \dfrac{1}{{n\choose i}} &= -\sum_{j=0}^{i}(-1)^{i-j}{i \choose j}\frac{i-j}{n-j} \\ &= -\sum_{j=0}^{i}(-1)^{i-j}{i \choose j}\frac{i-n}{n-j} - \sum_{j=0}^{i}(-1)^{i-j}{i \choose j}\frac{n-j}{n-j} \\ &= -\sum_{j=0}^{i}(-1)^{i-j}{i \choose j}\frac{i-n}{n-j} \\ \end{align} which gives the conclusion \begin{align} \dfrac{1}{{n\choose i}(n-i)} = \sum_{j=0}^{i}(-1)^{i-j}{i \choose j}\frac{1}{n-j} = \sum_{j=0}^{i}(-1)^{j}{i \choose j}\frac{1}{n-(i-j)} \end{align}

Clearly, if we do things in the inverse direction, we can prove the the partial fraction decomposition with $(a^n -b^n)= (a-b)(a^{n-1} + \cdots + b^{n-1})$.

Petite Etincelle
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    This is not a combinatorial proof as requested by the OP. – Semiclassical Aug 17 '14 at 13:43
  • @Semiclassical It uses combinatorial techniques, doesn't it? It also tells how to continue in the way OP started – Petite Etincelle Aug 17 '14 at 13:54
  • Not really: given the way the OP proceeds in their proof (by trying interpret the identity as two ways of counting some set of objects) it would appear to me that they mean 'combinatorial' argument in the spirit of this Wikipedia article. – Semiclassical Aug 17 '14 at 14:08
  • @Semiclassical What I see in OP's post is that OP looks for a combinatorial argument which proves the identity. Certainly there are more than one way to do that. OP tried one direction, but failed to go ahead. I showed how to go on in this direction by using only combinatorial arguments. Certainly the accepted answer is easier to read(at least for me). I just wanted to write another solution, which is not so natural, but succeeded in the started direction. – Petite Etincelle Aug 17 '14 at 14:20
  • Ok, I see your point. It'd be better to make that explicit in your answer, though (plus, I can't remove my downvote unless you edit your post.) – Semiclassical Aug 17 '14 at 14:25
  • @Semiclassical Keep it here, It's just some interpretation disagreement. I won't take it in the wrong way. Thank you for the discussion. – Petite Etincelle Aug 17 '14 at 14:44