Question

Suppose that a and b are real numbers with 0 < b < a . Prove that if n is a positive integer, then $a^n-b^n\le n{a}^{n-1}(a-b)$

Short answer

n is a positive integer, then $a^n-b^n\le n{a}^{n-1}(a-b)$

Step-by-step solution

Step: 1

If n=1,

$\begin{array}{r}{a}^1-b^1\le 1a^{1-1}(a-b)\\ (a-b)\le (a-b)\end{array}$

it is true for n=1.

Step: 2

Let P(k) be true.

$a^k-b^k\le k{a}^{k-1}(a-b)$

We need to prove that P(k) is true.

Step: 3

$\begin{array}{r}{a}^k-b^k\le (a-b)\sum _{i=0}^{k-1} a^{k-i-1}{b}^i\\ \le (a-b)\sum _{i=0}^{k-1} a^{k-i-1}{a}^i\\ =(a-b)\sum _{i=0}^{k-1} a^{k-i-1+i}\\ =(a-b)\sum _{i=0}^{k-1} a^{k+1}\\ =(a-b)k{a}^{k-1}\\ \le k{a}^{k-1}(a-b)\end{array}$

It is true for P(k) is true