The question is to prove $a^n – b^n \le na^{n – 1}(a-b)$ with induction where $0 < b < a$ and $ \forall n \ge 1$, $n \in \mathbb{Z}$
I finished the different parts of the proof, including the predicate, base case, and inductive step, but I want to make sure that I'm not missing anything – especially in the inductive step.
I'm also a little confused about moving from Step 1 to the rest of the steps because it seems that the inductive hypothesis applies to $b(a^n-b^n$, but I'm not sure how the other part of the expression, namely $a^n(a-b)$ is used.
Let $P(n) = a^n – b^n \le na^{n – 1}(a-b)$
Base case: $P(0) = a – b \le a – b$
Inductive step – need to prove $P(n + 1)$:
- $a^{n + 1} – b^{n + 1} = a^{n + 1} – a^nb + a^nb – b^{n + 1} = a^n(a-b) + b(a^n – b^n)$
- Because $0 < b < a$, $b(a^n-b^n) \le a(a^n – b^n)$
- Using the induction hypothesis, $a(a^n – b^n) \le a \cdot n \cdot a^{n – 1}(a-b)$
- $\rightarrow a(a^n – b^n) \le na^{n}(a-b) \le (n + 1)a^{n}(a-b)$
- $\rightarrow a(a^n – b^n) \le (n + 1)a^{n}(a-b)$
Best Answer
\begin{align} &a^{n+1} - b^{n+1} \\ =\ &a^{n+1} - a^nb + a^nb - b^{n+1} \tag{your step $1$} \\ =\ &a^n(a - b) + \color{red}{b(a^n - b^n)} \\ \leq\ &a^n(a - b) + \color{red}{a(a^n - b^n)} \\ \leq\ &a^n (a - b) + \color{red}{a \cdot na^{n-1}(a-b)} \\ \leq \ &a^n(a - b) + \color{red}{n\cdot a^n(a-b)} \tag{your step $4$}\\ =\ &(1 + n)a^n(a - b) \end{align}