Chapter 12.5, Problem 37E

a.

To determine

To Find: $p\left(n\right)=n^2-n+11$ is prime number for all $n$ is True or False.

Answer to Problem 37E

False

Explanation of Solution

Given information:

  $p\left(n\right)=n^2-n+11$ is Prime number For All $n$ .

Proof:

For $n=11$

We get

  $\begin{array}{l}p\left(n\right)=n^2-n+11\\ \Rightarrow p\left(11\right)={11}^2-11+11\\ \Rightarrow p\left(11\right)=121-11+11\\ \Rightarrow p\left(11\right)=121\end{array}$

As 121 is not a prime number

  $p\left(n\right)=n^2-n+11$ Is a prime number for all $n$ is False.

b.

To determine

To Find: $n^2>n$ for all $n\ge 2$ is True or False.

Answer to Problem 37E

True.

Explanation of Solution

Given information:

  $n^2>n$ for all $n\ge 2$

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 the natural numbers. The first step is to prove True for $n=2$ .

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:

Put $n=2$

We get

  $\begin{array}{l}{n}^2>n\\ \Rightarrow 2^2>2\\ \Rightarrow 4>2\end{array}$

It True for $n=2$ .

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

So

  $k^2>k$ is True ---------(1)

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

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

We need to prove equation-(2)

From equation-(1)

We get

  $k^2>k$

Adding with $2k+1$ on both sides

We get

  $\begin{array}{l}\Rightarrow k^2+2k+1>k+2k+1\\ \Rightarrow {\left(k+1\right)}^2>3k+1\end{array}$

We know

  $3k+1>k+1$ for all $k\ge 2$

So

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

Which is equation-(2)

So

  $n^2\ge n$ for $n\ge 2$ .

Which is True for $n=k+1$

And hence the proof by Mathematical induction.

c.

To determine

To Find: $2^{2n+1}+1$ is divisible by 3 for all $n\ge 1$ is True or False.

Answer to Problem 37E

True

Explanation of Solution

Given information:

  $2^{2n+1}+1$ is divisible by 3 for all $n\ge 1$

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 the natural numbers. The first step is to prove True for $n=1$ .

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:

Given formula is

  $2^{2n+1}+1$ is divisible by 3 for all $n\ge 1$

Put $n=1$

We get

  $\begin{array}{l}{2}^{2n+1}+1\\ =2^{2\left(1\right)+1}+1\\ =2^{2+1}+1\\ =2^3+1\\ =8+1\\ =9\end{array}$

It True for $n=1$ .

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

So

  $2^{2\left(k\right)+1}+1$ for $k\ge 1$ is True ---------(1)

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

  $\begin{array}{l}{2}^{2n+1}+1\\ =2^{2\left(k+1\right)+1}+1\\ =2^{2k+2+1}+1\\ =2^{2k+1}\cdot 2^2+1\\ =4\cdot 2^{2k+1}+4-3\\ =4\left(2^{2k+1}+1\right)-3\end{array}$

As

  $2^{2k+1}+1$ is divisible 3 and 3 also divisible by 3

So, $4\left(2^{2k+1}+1\right)-3$ is also divisible by 3.

  $2^{2n+1}+1$ is divisible for $n\ge 1$ .

Which is True for $n=k+1$

And hence the proof by Mathematical induction.

d.

To determine

To Find: $n^3\ge {\left(n+1\right)}^2$ for all $n\ge 2$ is True or False.

Answer to Problem 37E

False

Explanation of Solution

Given information:

  $n^3\ge {\left(n+1\right)}^2$ for all $n\ge 2$ .

Proof:

Given formula is

  $n^3\ge {\left(n+1\right)}^2$ for all $n\ge 2$

Put $n=2$

We get

  $\begin{array}{l}{n}^3\ge {\left(n+1\right)}^2\\ \Rightarrow 2^3\ge {\left(2+1\right)}^2\\ \Rightarrow 8\ge {\left(3\right)}^2\\ \Rightarrow 8\ge 9\end{array}$

Which is False

So,

  $n^3\ge {\left(n+1\right)}^2$ for all $n\ge 2$ is False.

e.

To determine

To Find: $n^3-n$ is divisible by 3 for all $n\ge 2$ is True or False.

Answer to Problem 37E

True

Explanation of Solution

Given information:

  $n^3-n$ is divisible by 3 for all $n\ge 2$ .

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 the natural numbers. The first step is to prove True for $n=2$ .

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:

Given formula is

  $n^3-n$ is divisible by 3 for all $n\ge 2$

Put $n=2$

We get

  $\begin{array}{l}{n}^3-n\\ =2^3-2\\ =8-2\\ =6\end{array}$

As 6 is divisible by 3

It True for $n=2$ .

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

So

  $k^3-k$ is divisible by 3 is True ---------(1)

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

  $\begin{array}{l}{n}^3-n\\ ={\left(k+1\right)}^3-\left(k+1\right)\\ =k^3+1+3k^2+3k-k-1\\ =\left(k^3-k\right)+3\left(k^2+k\right)\end{array}$

As

  $k^3-k$ is divisible 3 and $3\left(k^2+k\right)$ also divisible by 3

So, $\left(k^3-k\right)+3\left(k^2+k\right)$ is also divisible by 3.

  $n^3-n$ is divisible by 3 for $n\ge 2$ .

Which is True for $n=k+1$

And hence the proof by Mathematical induction.

f.

To determine

To Find: $n^3-6n^2+11n$ is divisible by 6 for all $n\ge 1$ is True or False.

Answer to Problem 37E

True

Explanation of Solution

Given information:

  $n^3-6n^2+11n$ is divisible by 6 for all $n\ge 1$ .

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 the natural numbers. The first step is to prove True for $n=1$ .

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:

Given formula is

  $n^3-6n^2+11n$ is divisible by 6 for all $n\ge 1$

Put $n=1$

We get

  $\begin{array}{l}{n}^3-6n^2+11n\\ =1^3-6{\left(1\right)}^2+11\left(1\right)\\ =1-6+11\\ =12-6\\ =6\end{array}$

As 6 is divisible by 6

It True for $n=1$ .

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

So

  $k^3-6k^2+11k$ for all $k\ge 1$ is divisible by 6 is True ---------(1)

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

  $\begin{array}{l}{n}^3-6n^2+11n\\ ={\left(k+1\right)}^3-6{\left(k+1\right)}^2+11\left(k+1\right)\\ =k^3+3k^2+3k+1-6\left(k^2+2k+1\right)+11k+11\\ =k^3+3k^2+3k+1-6k^2-12k-6+11k+11\\ =\left(k^3-6k^2+11k\right)+3k^2+3k+1-12k-6+11\\ =\left(k^3-6k^2+11k\right)+3k^2+3k-12k+6\\ =\left(k^3-6k^2+11k\right)+3k\left(k+1\right)-6\left(2k-1\right)\end{array}$

As

  $\left(k^3-6k^2+11k\right)$ is divisible 6 and $6\left(2k-1\right)$ also divisible by 6

  $3k\left(k+1\right)$ is also multiple of 6 because $k\left(k+1\right)$ is multiple of 2 for all values of k

So $3k\left(k+1\right)$ is also divisible of 6.

So, $\left(k^3-6k^2+11k\right)+3k\left(k+1\right)-6\left(2k-1\right)$ is also divisible by 3.

  $n^3-6n^2+11n$ is divisible by 6 for all $n\ge 1$ .

Which is True for $n=k+1$

And hence the proof by Mathematical induction.