0

Prove by induction: $b^n-a^n=(b-a)\cdot \sum_{k=0}^{n-1}{a^k\cdot b^{n-1-k}}$ $\forall n\in \mathbb{N}$ and $a,b\in \mathbb{R}$

$\exists n\in \mathbb{N}:b^n-a^n=(b-a)\cdot \sum_{k=0}^{n-1}{a^k\cdot b^{n-1-k}}$

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

Base case: $n_0=1: b^1-a^2=(b-a)\cdot \underbrace{\sum_{k=0}^{1-1}{a^k\cdot b^{1-1-k}}}_{=1} \quad \checkmark$

Inductive step:

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

The sigma sign is what stresses me out, how can I deal with it?

José Carlos Santos
  • 427,504

3 Answers3

1

$$\sum_{k=0}^na^kb^{n-k}=a^n+b\sum_{k=0}^{n-1}a^kb^{n-k-1}=a^n+b\frac{b^n-a^n}{b-a}=\frac{b^{n+1}-a^{n+1}}{b-a}$$

ajotatxe
  • 65,084
1

Inductive step

$$\begin{align*}b^{n+1}-a^{n+1}&=(b-a)\cdot \sum^n_{k=0}a^kb^{n-k}\\&=(b-a)\cdot \bigg(a^n+b\sum^{n-1}_{k=0}a^kb^{n-k-1}\bigg)\\&=a^n(b-a)+b\cdot(b^n-a^n)\\&=b^{n+1}-a^{n+1}\end{align*}$$

Community
  • 1
Dr. Mathva
  • 8,621
  • .... how did I miss that! :-) –  May 07 '19 at 17:30
  • Shouldn't it be $(b-a)\cdot \left[(a^n+b)\sum_{k=0}^{n-1}{a^k\cdot b^{n-1-k}}\right]$ in the second line? –  May 07 '19 at 18:03
  • Not really, observe that \begin{align}\sum^n_{k=0}a^kb^{n-k}&=b^n+ab^{n-1}+a^2b^{n-2}+\ldots+a^{n-1}b+a^n\&=b\cdot (b^{n-1}+ab^{n-2}+a^2b^{n-3}+\ldots +a^{n-1})+a^n\&=b\cdot \sum^{n-1}_{k=0}a^kb^{n-k-1}+a^n \end{align} – Dr. Mathva May 07 '19 at 18:08
0

You might want to note that, by your induction hypothesis,

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

Then when you do the induction, you will be verifying

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

You can substitute in $b^n - a^n$ for the summation by your hypothesis.

PrincessEev
  • 43,815