Chapter 12.5, Problem 24E

To determine

To Prove: $100n\le n^2$ for all the natural numbers $n\ge 100$ .

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 $n=100$ .

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 $n=100$ .

Given formula is

  $100n<n^2$ for all natural numbers $n\ge 100$ .

Put $n=100$

We get

  $\begin{array}{l}100n\le n^2\\ \Rightarrow 100\left(100\right)\le {100}^2\\ \Rightarrow 10000\le 10000\end{array}$

It True for $n=100$ .

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

So

  $100k\le k^2$ is True ---------(1)

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

  $100\left(k+1\right)\le {\left(k+1\right)}^2$ ------(2)

We need to prove equation-(2)

From equation-(1)

We get

  $100k\le k^2$

Adding with $2k+1$ on both sides

We get

  $\Rightarrow 100k+2k+1\le k^2+2k+1$

  $\Rightarrow 100k+\left(2k+1\right)\le {\left(k+1\right)}^2$

  $100<2k+1$ for $k\ge 100$

  $\Rightarrow 100k+100\le {\left(k+1\right)}^2$

  $\Rightarrow 100\left(k+1\right)\le {\left(k+1\right)}^2$

Which is equation-(2)

So

  $100n\le n^2$

Which is True for $n=k+1$

And hence the proof by Mathematical induction.