Chapter 12.5, Problem 23E

To determine

To Prove: ${\left(1+x\right)}^n\ge 1+nx$ ,if $x>-1$ and for all natural numbers.

Explanation of Solution

Given information:

Use Mathematical induction to prove the above mentioned formula.

Concept used:

1.By using Mathematical induction technique we need to prove that this formula holds True for all natural numbers. The first step is to prove True for base value of $n$ that is $n=1$ .

2.Second step by using induction hypothesis of mathematical induction we assume for $n=k$ given formula is True.

3.Third step is we need to prove that above formula is True for $n=k+1$ by using formula is True for $n=k$ .

Proof:

First, we need to prove the above formula is True for $n=1$ .

Given formula is

  ${\left(1+x\right)}^n\ge 1+nx$

Put $n=1$

We get

  $\begin{array}{l}{\left(1+x\right)}^n\ge 1+nx\\ \Rightarrow {\left(1+x\right)}^1\ge 1+\left(1\right)x\\ \Rightarrow 1+x\ge 1+x\end{array}$

It True for $n=1$ .

Let us assume it is True for $n=k$

So

  ${\left(1+x\right)}^k\ge 1+kx$ for $x>-1$ is True ---------(1)

We need to prove it is True for $n=k+1$

  ${\left(1+x\right)}^{k+1}\ge 1+\left(k+1\right)x$

Taking Left hand side

We get

Left hand side = ${\left(1+x\right)}^{k+1}$

  $\Rightarrow$ Left hand side = ${\left(1+x\right)}^k\left(1+x\right)$

From equation− (1) ${\left(1+x\right)}^k\ge 1+kx$

We get

  $\Rightarrow$ Left hand side $\ge \left(1+kx\right)\left(1+x\right)$

  $\Rightarrow$ Left hand side $\ge 1+kx+x+k{x}^2$

  $\Rightarrow$ Left hand side $\ge \left(1+\left(k+1\right)x\right)+k{x}^2$

Since $k{x}^2\ge 0$

  $\Rightarrow$ Left hand side $\ge \left(1+\left(k+1\right)x\right)$

  $\Rightarrow$ Left hand side $\ge$ Right hand side.

So

  ${\left(1+x\right)}^n\ge 1+nx$ , $x>-1$ is True for $n=k+1$

And hence the proof by Mathematical induction.