Question

3^(2n+2)-8n-9 is divididable
by 64​

Answer 1

Answer:

Let 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+64k+64

=64(9m+k+1),   Which is divisibility by 64

Thus p(k+1)is true whenever p(k) is true.

Hence, by principal mathematical induction,

p(x) is true for all natural number p(x)=3  

2n+2

−8x−9 is divisible by 64 n∈N

Step-by-step explanation:

Answer 2

your answer.....

If it helps you mark me as brainlist......