Question

prove by principal of mathematical induction
p(n)=9^n-1 divisible by 8 where n is a natural number?​​

Answer 1

Answer:

The principal of mathematical induction

The statement is true for n=1,.e,

•lf the statement is true for n = kwhere k is a positive interger, then the statement also true for all cases of.n=k+1 i, ex(k) leads to the truth of X (k+1).hope it helps

Answer 2

Step-by-step explanation:

p(x)=3

2n+2

−8x−9 is divisible by 64 …..(1)

When put n=1,

p(1)=3

4

−8−9=64 which is divisible by 64

Let n=k and we get

p(k)=3

2k+2

−8k−9 is divisible by 64

3

2k+2

−8k−9=64m where m∈N …..(2)

Now we shall prove that p(k+1) is also true

p(k+1)=3

2(k+1)+2

−8(k+1)−9 is divisible by 64.

Now,

p(k+1)=3

2(k+1)+2

−8(k+1)−9=3

2

.3

2k+2

−8k−17

=9.3

2k+2

−8k−17

=9(64m+8k+9)−8k−17

=9.64m+72k+81−8k−17

=9.64m+