**To Prove**

3^{n} – 1 is a multiple of 2 ∀ n ∈ N

**Proof**

We can prove the given statement using many ways. We will discuss 2 of them.

**Direct Proof**

We know the product of two odd numbers is always an odd number. This implies that the term 3^{n} is always odd ∀ n ∈ N. On subtracting an odd number from another odd number, we get an even number.

This implies that the term **3 ^{n} – 1** is even since both 3

^{n}and 1 are odd numbers. Therefore, we can say that 3

^{n}– 1 is a multiple of 2 since all even numbers are multiples of 2. Hence Proved.

**Mathematical Induction**

**Given**

3^{n} – 1 is a multiple of 2 ∀ n ∈ N

**STEP 1**

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

3^{n} – 1 = 3^{1} – 1 = 3 – 1 = 2

The statement is true for n = 1.

**STEP 2**

Let the statement be true for n = k. 3^{k} – 1 is a multiple of 2. We can write it as

3^{k} – 1 = 2x, ∀ x ∈ N ……..(1)

**STEP 3**

We need to check if the statement is true for n = k + 1 given that the statement is true for n = k.

………(2)

Substituting equation (1) in (2)

which is a multiple of 2.

Thus, the statement is true for n = k + 1 given that it is true for n = k. Therefore, by the principle of mathematical induction, the given statement “*3 ^{n} – 1 is a multiple of 2*” is true for all n ∈ N.