**To Prove**

4^{n} – 3n – 1 is divisible by 9 ∀ n ∈ N

**Proof**

We will prove the given statement using mathematical induction.

**Step 1**

First, we will check if the given statement is true for n = 1.

4^{n} – 3n – 1 = 4^{1} – 3(1) – 1 = 4 – 3 – 1 = 0

The statement is true for n = 1 since 0 is divisible by 9.

**Step 2**

In this step, we will assume that the statement is true for n = k. This implies that 4^{k} – 3k – 1 is divisible by 9. We can say

4^{k} – 3k – 1 = 9x, ∀ x ∈ N ……(1)

**Step 3**

In this step, we will check if the statement is true for n = k + 1 given that it is true for n = k.

Substituting equation (1) in (2)

9(4x+k) is divisible by 9. This implies that the given statement is true for n = k + 1 given that the statement is true for n = k. Therefore, by the principle of mathematical induction, the given statement “4^{n} – 3n – 1 is divisible by 9″ is true for all n ∈ N.