Prove the following statement by mathematical induction.For every integer

n0, 7ⁿ1

is divisible by 6.Proof (by mathematical induction): LetP(n) be the following sentence.

7ⁿ1

is divisible by 6.We will show thatP(n) is true for every integer

n0.

ShowthatP(0)istrue:

SelectP(0) from the choices below.

(7⁰1) | 6

6 | (7⁰1)

1 is a factor of 7⁰1

6 is a multiple of 7⁰1

The truth of the selected statement follows from the definition of divisibility and the fact that

7⁰1 = .

Showthatforeachintegerk0,ifP(k)istrue,thenP(k+ 1)istrue:

Letkbe any integer with

k0,

and suppose that

P(k)

is true. Select

P(k)

from the choices below.

6 is a multiple of 7^k1

6 is divisible by (7^k1)

1 is a factor of 7^k1

(7^k1) is divisible by 6

[This is

P(k),

the inductive hypothesis.]We must show that

P(k+ 1)

is true. Select

P(k+ 1)

from the choices below.

6 is a multiple of 7^{k+ 1}1

6 is divisible by (7^{k+ 1}1)

1 is a factor of 7^{k+ 1}1

(7^{k+ 1}1) is divisible by 6

By the inductive hypothesis and the definition of divisibility, there exists an integerrsuch that

7^k1 = 6r,

and so

7^k= 6r+ 1.

Now

7^k+11 = 7^k71.

When

6r+ 1

is substituted for

7^k

in the above equation, and the right-hand side is simplified, the result can be expressed in terms ofkandras follows.

7^k+11 = 6

This quantity is an integer becausekandrare integers.Select the final sentence from the choices below.

Hence, (7^k+11) is divisible by 6, and soP(k+1) is true, which completes the inductive step.

Hence, (7^k+11) is divisible by 6, and soP(k+1) is false, which completes the inductive step.

Hence, 6 is a multiple of (7^k+11), and soP(k+1) is false, which completes the inductive step.

Hence, 1 is a factor of (7^k+11), and soP(k+1) is false, which completes the inductive step.

Hence, 1 is a factor of (7^k+11), and soP(k+1) is true, which completes the inductive step.

Hence, 6 is divisible by (7^k+11), and soP(k+1) is true, which completes the inductive step.

[Thus both the basis and the inductive steps have been proven,and so the proof by mathematical induction is complete.]