Chapter 9.4, Problem 30E

To determine

To prove the inequality ${\left(1+a\right)}^n\ge na$ for all integers $n\ge 1$ and $a>0$ by using mathematical induction.

Answer to Problem 30E

The inequality ${\left(1+a\right)}^n\ge na$ for all integers $n\ge 1$ and $a>0$ is valid.

Explanation of Solution

Given:

Let $P_n:{\left(1+a\right)}^n\ge na$ for all integers $n\ge 1$ and $a>0$ .

Concept used:

The Principle of Mathematical Induction states that for a statement $P_n$ involving the positive integer n . If

1. $P_1$ is true, and

2. for every positive integer k , the truth of $P_k$ implies the truth of $P_{k+1}$ then the statement $P_n$ must be true for all positive integers n .

Calculation:

First to show that the property is valid for $n=1$ . When $n=1$ , the property is valid because:

  $\begin{array}{l}{P}_1:{\left(1+a\right)}^1\ge \left(1\right)a\\ 1+a\ge a\end{array}$

Now, let us assume that for $n=k$ ,

  $P_k:{\left(1+a\right)}^k\ge ka$ ,

we need to show that for $n=k+1$ ,

  $P_{k+1}:{\left(1+a\right)}^{k+1}\ge \left(k+1\right)a$ .

For $n=k+1$ ,

  ${\left(1+a\right)}^{k+1}={\left(1+a\right)}^k\left(1+a\right)$

From (1) and the assumption taken we know, $\left(1+a\right)\ge a$ and ${\left(1+a\right)}^k\ge ka$ , therefore,

  $\begin{array}{c}{\left(1+a\right)}^{k+1}={\left(1+a\right)}^k\left(1+a\right)\\ \ge ka+a\\ =\left(k+1\right)a\end{array}$

Hence, proved.

Therefore, by combining all the results, we can conclude by the mathematical induction that the inequality is valid for all $n\ge 1$ .