Question

Prove that 3^2n+2-8n-9 is divisible by 8 for all n>1.​

Answer 1

Answer:

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Step-by-step explanation:

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