Question

Use mathematical induction to prove the following statement: 8"-1 is a multiple of 7 for all neN

Answer 1

Answer: with Step-by-step explanation:

We are given that

P(n)=$8^n-1$

We have to prove that given statement is a multiple of 7 using mathematical induction for all natural numbers n belongs to N.

Suppose n=1

Then P(1)=8-1=7

Hence, it is a multiple of 7 .Therefore, it is true for n=1

We suppose that it is true for n=k

Then P(k)=$8^k-1$ is a multiple of 7.

We shall prove that it is true for n=k+1

P(k+1)=$8^{k+1}-1$ is a multiple of 7

LHS=$8^{k+1}-1$

=$8^k\cdot8-1$

=$8^k\cdot8-8+8-1$

=$8(8^k-1)+7$

=$8\cdot 7a+7$  because $8^k-1$ is a multiple of 7 therefore$8^k-1=7a$

=$7(8a+1)$

P(k+1) is a multiple of 7.

Therefore, P(n) is true for all natural numbers belongs to N.

   

Answer 2

Answer:

Step-by-step explanation:

To prove that $8^n-1$ is a multiple of 7.

Proof by induction:

Let n =1.  Then we have 8-1 =7 is a multiple of 7

Thus the P(1) is true.  

Let us assume that P(n) is true.

$8^n-1$ = 7l is true for some integer l

To check for P(n+1)

$8^{n+1} -1$=LHS

=$8^n(8)-1\\=(7l+1)8-1\\=56l+7\\=7(8l+1)$

Thus we find that this is also a multiple of 7.

If true for n, then true for n+1

Hence we find that since true for 1, we have it is true for all natural numbers.