TITLE: 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 [0 upvotes]: 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?

REPLY [1 votes]: Inductive step
$$\begin{align*}b^{n+1}-a^{n+1}&=(b-a)\cdot \sum^n_{k=0}a^kb^{n-k}\\&=(b-a)\cdot \bigg(a^n+b\sum^{n-1}_{k=0}a^kb^{n-k-1}\bigg)\\&=a^n(b-a)+b\cdot(b^n-a^n)\\&=b^{n+1}-a^{n+1}\end{align*}$$
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