Prove that 3^n-1 is a multiple of 2

To Prove

3n – 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 3n is always odd ∀ n ∈ N. On subtracting an odd number from another odd number, we get an even number.

This implies that the term 3n – 1 is even since both 3n and 1 are odd numbers. Therefore, we can say that 3n – 1 is a multiple of 2 since all even numbers are multiples of 2. Hence Proved.

Mathematical Induction

Given

3n – 1 is a multiple of 2 ∀ n ∈ N

STEP 1

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

3n – 1 = 31 – 1 = 3 – 1 = 2

The statement is true for n = 1.

STEP 2

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

3k – 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.

3^{k+1}-1 = 3^k.3-1 ………(2)

Substituting equation (1) in (2)

\\*3^{k+1}-1 = (2x+1).3 - 1\\*3^{k+1}-1 = 6x+3-1\\*3^{k+1}-1 = 6x + 2\\*3^{k+1}-1 = 2(3x + 1)

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 “3n – 1 is a multiple of 2” is true for all n ∈ N.

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