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 $a,b\in \mathbb{R}$

Question

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?

Answer 1

Inductive step

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

Answer 2

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.

Answer 3

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