Help with an arithmetic proof by induction

[Entertainment Unit]

Homework Statement

Show that if $0 \leq a < b$, then
$$\frac {b^{n + 1} - a^{n + 1}} {b - a} < (n + 1)b^n$$

Homework Equations

None that I'm aware of.

The Attempt at a Solution

Proof (Induction)
1. Basis Case: Suppose $n = 1$. It follows that:
$$\frac {b^{1 + 1} - a^{1 + 1}} {b - a} < (1 + 1)b^1$$
$$\frac {b^2 - a^2} {b - a} < 2b$$
$$b^2 - a^2< 2b(b - a)$$
$$b^2 - a^2< 2b^2- 2ab$$
$$-a^2< 2b^2$$

which is true for all $0 \leq a < b$.

2. Inductive Step. Let $k = n \geq 1$ and suppose that:
$$\frac {b^{k + 1} - a^{k + 1}} {b - a} < (k + 1)b^k$$
This is where I can't make any progress. I can get the right-hand side to read $(k + 2)b^{k + 1}$ but doing so leaves the left-hand side of the inequality in a state that I can't get it to $\frac {b^{k + 2} - a^{k + 2}} {b - a}$.

 

[pasmith]

You don't need to use induction here.
The relevant equation you are missing is <br /> b^{n+1} - a^{n+1} = (b-a)(b^n + ab^{n-1} + \dots + a^{n-1}b + a^n)<br />

 

[Entertainment Unit]

Thank you! I went down the path of proving it directly but struggled with that too.

Here's my completed proof:

Proof: Suppose

$$\frac {b^{n + 1} - a^{n + 1}} {b - a} < (n+1)b^n$$

It follows that
$$\frac {(b - a)(b^n + ab^{n - 1} + \dots + a^{n - 1}b + a^n)} {b - a} < (n+1)b^n$$
$$b^n + ab^{n - 1} + \dots + a^{n - 1}b + a^n < (n+1)b^n$$
$$b^n < (n+1)b^n$$

which is true for $b \gt 0, n \geq 1$. Therefore,

$$\frac {b^{n + 1} - a^{n + 1}} {b - a} < (n+1)b^n\text{ ■}$$

 

[pasmith]

The proof is even easier: It follows from 0 \leq a &lt; b that b^{n-k} a^{k} &lt; b^n for 1 \leq k \leq n. Hence
<br /> \frac{b^{n+1} - a^{n+1}}{b-a} = \sum_{k=0}^n b^{n-k}a^k &lt; \sum_{k=0}^n b^n = (n+1)b^n.<br />

 

[Entertainment Unit]

The proof is even easier: It follows ...

Thanks for this!