32n + 2 – 8n – 9 is divisible by 8 for all n ϵ N

Asked by Sakshi | 1 year ago |  129

##### Solution :-

Let P (n): 32n + 2 – 8n – 9 is divisible by 8

Let us check for n = 1,

P (1): 32.1 + 2 – 8.1 – 9

: 81 – 17

: 64

P (n) is true for n = 1. Where, P (n) is divisible by 8

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.

P (k): 32k + 2 – 8k – 9 is divisible by 8

: 32k + 2 – 8k – 9 = 8λ … (i)

We have to prove,

32k + 4 – 8(k + 1) – 9 is divisible by 8

3(2k + 2) + 2 – 8(k + 1) – 9 = 8μ

So,

= 32(k + 1).32 – 8(k + 1) – 9

= (8λ + 8k + 9)9 – 8k – 8 – 9

= 72λ + 72k + 81 – 8k – 17 using equation (1)

= 72λ + 64k + 64

= 8(9λ + 8k + 8)

= 8μ

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.

Answered by Aaryan | 1 year ago

### Related Questions

#### Given an example of a statement P (n) such that it is true for all n ϵ N.

Given an example of a statement P (n) such that it is true for all n ϵ N.

#### a + (a + d) + (a + 2d) + … + (a + (n-1)d) = n/2 [2a + (n-1)d]

a + (a + d) + (a + 2d) + … + (a + (n-1)d) = $$\dfrac{n}{2}$$ [2a + (n-1)d]

#### 72n + 23n – 3. 3n – 1 is divisible by 25 for all n ϵ N

72n + 23n – 3. 3n – 1 is divisible by 25 for all n ϵ N