Chapter 12.4, Problem 27AYU

In Problems 23-27, prove each statement.

${\left(1\text{ + }a\right)}^n\ge \text{ 1 + }na$, for $a\text{ > 0}$

To determine

To prove: The given statement ${\left(1 + a\right)}^n \ge 1 + na,$ for $a > 0$ is true for all natural numbers $n$ using the Principle of Mathematical Induction.

Answer to Problem 27AYU

As the statement is true for the natural number $\left(k + 1\right)$ terms, it is true for all natural numbers by the theorem of mathematical induction.

Explanation of Solution

Given:

Statements says the series ${\left(1 + a\right)}^n \ge 1 + na,$ for $a > 0$ is true for all natural number.

Formula used:

The Principle of Mathematical Induction

Suppose that the following two conditions are satisfied with regard to a statement about natural numbers:

CONDITION I: The statement is true for the natural number 1.

CONDITION II: If the statement is true for some natural number $k$, it is also true for the next natural number $k + 1$. Then the statement is true for all natural numbers.

Proof:

Consider the statement

${\left(1 + a\right)}^n \ge 1 + na$ for $a > 0$    -----(1)

Step 1: Show that statement (1) is true for the natural number $n = 1$.

That is ${\left(1 + a\right)}^1 = 1 + a = 1 + 1.a$. Hence the statement is true for the natural number $n = 1$.

Step 2: Assume that the statement is true for some natural number $k$.

That is ${\left(1 + a\right)}^k \ge 1 + ka$ for $a > 0$    -----(2)

Step 3: Prove that the statement is true for the next natural number $k + 1$.

That is, to prove that ${\left(1 + a\right)}^{k + 1} \ge 1 + \left(k + 1\right)a$, for $a > 0$

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

$\ge \left(1 + ka\right)\left(1 + a\right)$

$= 1 + a + ka + k{a}^2$

$= 1 + \left(k + 1\right)a + k{a}^2$

$\ge 1 + ka\text{ for }a > 0$

As the statement is true for the natural number $\left(k + 1\right)$ terms, it is true for all natural numbers by the theorem of mathematical induction.