Different proofs of $\,a^n-b^n =(a-b)\sum_{i=0}^{n-1} a^i b^{n-1-i} $?

Question

How many different proofs are there that $a^n-b^n =(a-b)\sum_{i=0}^{n-1} a^i b^{n-1-i} $ for positive integer $n$ and real $a, b$?

You can use any techniques you want. My proof just uses algebra, summation, and induction, but if you want to use invariant sheaves over covalent topologies, that is fine.

I decided that I would try to produce a proof by induction. I find it interesting that my proof shows that if it is true for $n-1$, then it is true for $n+1$. This means that two base cases have to be proven: $n=1$ and $n=2$. Fortunately, those are easy.

I am sure that my proof is known, but I do not recall having seen it before.

Here is the induction step:

$\begin{array}\\ a^{n+1}-b^{n+1} &=a^{n+1}-a^nb+a^nb-b^{n+1}\\ &=a^{n+1}-a^nb +a^nb-ab^n +ab^n-b^{n+1}\\ &=(a-b)a^n +ab(a^{n-1}-b^{n-1}) +(a-b)b^n\\ &=(a-b)(a^n+b^n) +ab(a^{n-1}-b^{n-1}) \\ &=(a-b)(a^n+b^n) +ab((a-b)\sum_{i=0}^{n-2} a^i b^{n-2-i}) \ \ \text{(The induction hypothesis)} \\ &=(a-b)(a^n+b^n+ab\sum_{i=0}^{n-2} a^i b^{n-2-i}) \\ &=(a-b)(a^n+b^n+\sum_{i=0}^{n-2} a^{i+1} b^{n-1-i}) \\ &=(a-b)(a^n+b^n+\sum_{i=1}^{n-1} a^{i} b^{n-i}) \\ &=(a-b)\sum_{i=0}^{n} a^{i} b^{n-i} \\ \end{array} $

Answer 1

By telescopy $\ f_n = x^n\,\Rightarrow \displaystyle \overbrace{f_n-f_0 =\sum_{k\,=\,0}^{n-1}\left[f_{k+1}\:\!-\:\!f_k\right]}^{\textstyle\ \ x^{n} - 1 \:\!=\:\! \sum\ {\overbrace{[x^{k+1}-{x^k}^{\phantom |}\!}^{\Large \color{#0a0}{(x\,-\,1)}\,\color{#c00}{x^k}\!\!}]}_{\phantom{|_|}}} =\, \color{#0a0}{(x\!-\!1)}\sum_{k\,=\,0}^{n-1}\, \color{#c00}{x^k} $

The sought result (Factor Theorem) follows by homogenization, i.e. $\, x\to a/b\,$ then scale by $\,b^n.$

Remark $\ $ The simple theorem employed to evaluate the above telescopic sum may be viewed as a discrete analog of the Fundamental Theorem of Integral Calculus

$$\begin{eqnarray} f &=& \sum \Delta f,\quad \Delta f(n) = f(n+1) -f(n)\\ f & =& \ \int D f,\quad D\, f(x) = f'(x)\end{eqnarray}$$

However, the proof of the discrete analog (the Fundamental Theorem of Difference Calculus) is just a trivial one-line induction exploiting telescopic cancellation.

Answer 2

Using $$\sum_{k=0}^{n-1}x^k={1-x^n\over1-x}$$ we get \begin{align} a^n-b^n &=a^n\left[1-\left({b \over a}\right)^n\right]\\ &=a^n\left[1-{b \over a}\right]\sum^{n-1}_{k=0}{b^k\over a^k}\\ &=\sum^{n-1}_{k=0}a^{n-k}b^k-\sum^{n-1}_{k=0}a^{n-k-1}b^{k+1}\\ &=(a-b)\sum^{n-1}_{k=0}a^{n-k-1}b^k \end{align}

Answer 3

$P(x)=x^n-1$ has the root $x=1$ so it factors as $P(x) = (x-1)Q(x)$. Differentiating the product $k \lt n$ times with the Leibniz rule and using that $(x-1)^{(j)} = 0$ for $j \ge 2\,$:

$$ n(n-1)\dots(n-k+1)\,x^{n-k} = (x-1)Q^{(k)}(x) + \binom{k}{1} Q^{(k-1)}(x) $$

Setting $\,x=0\,$ gives $\,Q^{k}(0)=k\,Q^{(k-1)}(0)\,$, so by induction $\,Q^{(k)}(0) = k!\,$, then by Taylor's theorem:

$$ Q(x) = \sum_{k=0}^{n-1} \frac{Q^{(k)}(0)}{k!}\,x^k = \sum_{k=0}^{n-1} \,x^k \quad\iff\quad x^n-1 = (x-1)\left(1 + x + \dots+x^{n-1}\right) $$

Homogenizing the latter equality produces the title identity.