Question

Using the principle of induction, prove that 64 is a factor of 3^(2n+2)- 8n-9 ∀ n∈N

Answer 1

Let p(n)=3$^{2n+2}$−8n−9 is divisible by 64 …..(1)
When put n=1,p(1)=34−8−9=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)=$3^{2(k+1)+2}−8(k+1)−9$ =$3^23^{2k+2}−8(k+1)−9$ = $9(3^{2k+2})−8(k+1)−9$ =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.