1

I'm working on an inductive proof, and I came across this line but I'm not sure how these are equivalent. Here is the link if more detail is needed. It's in box 2, first line.

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

cmk
  • 12,303
financial_physician
  • 769

2 Answers2

3

Expand the right hand side: $$\begin{align}(a-b)\sum_{i=0}^{k-1}a^{k-i-1}b^i&=a\sum_{i=0}^{k-1}a^{k-i-1}b^i-b\sum_{i=0}^{k-1}a^{k-i-1}b^i\\ &=\sum_{i=0}^{k-1}a^{k-i}b^i-\sum_{i=0}^{k-1}a^{k-i-1}b^{i+1}\\ &=\sum_{i=0}^{k-1}a^{k-i}b^i-\sum_{i=1}^{k}a^{k-i}b^{i}\\ &=a^k+\sum_{i=1}^{k-1}a^{k-i}b^i-b^k-\sum_{i=1}^{k-1}a^{k-i}b^{i}\\ &=a^k-b^k.\end{align}$$

Hagen von Eitzen
  • 374,180
1

The other way (left to right):

  • First prove by induction the special case: $$1-x^k=(1-x)\bigl(1+x+x^2+\dots+x^{k-1}\bigr).$$ The initial case ($k=1$) reduces to $\; 1-x=1-x$.

    Suppose now the formula is established for some $k\ge 1$, and consider

So the inductive step is proved.

  • Generalisation: we may suppose $a\ne 0$, and set $x=\dfrac ba$. Note that fot each $i$, $\;a^ix^i = b^i$, so

\begin{align} a^k-b^k=a^k\bigl(1-x^k\bigr)&=a^k(1-x)\bigl(1+x+x^2+\dots+x^{k-1}\bigr)\\ &=a(1-x)\,a^{k-1}\bigl(1+x+x^2+\dots+x^{k-1}\bigr) \\ &=(a-b)\bigl(a^{k-1}+a^{k-2}(ax)+a^{k-3}(a^2x^2)+\dots+a^{k-1}x^{k-1}\bigr)\\ &=(a-b)\bigl(a^{k-1}+a^{k-2}b+a^{k-3}b^2+\dots+b^{k-1}\bigr).\\ \end{align}

Bernard
  • 175,478
  • @OP The transformation used above is called (de-) homogenization. $\ \ $ – Bill Dubuque Jun 15 '19 at 20:26
  • I know. I didn't want to use it in this context. I used to give this explicit induction in the special case, as the simplest, non totally trivial, example to give to students who are introduced to this technique for the first time. – Bernard Jun 15 '19 at 20:32
  • In case you didn't know, "OP" means Original Poster, i.e. the question author is the target of my comment. Most students at this level haven't yet learned this simple idea. – Bill Dubuque Jun 15 '19 at 20:37