**Given**

2^{n} + 1 is divisible by 3 ∀ n ∈ O^{+}

**Proof**

We will prove the given statement by induction.

**STEP 1**

n = 1

2^{n} + 1 = 2^{1} + 1 = 3

3 is divisible by 3. Therefore, the statement is true for n = 1

**STEP 2**

Let the given statement be true for n = k

2^{k} + 1 = 3x

Now, we need to prove that if the statement is true for n = k then it is also true for n = k + 2

*Note: We are going to prove the statement to be true for n = k + 2 and not for n = k + 1 because k is odd. Therefore, k + 1 is even. The next odd number after k is k + 2.*

2^{k+2} + 1 = 2^{k}*4 + 1

= ( 3x – 1 )*4 + 1

= 12x – 4 + 1

= 12x – 3

= 3 * ( 4x – 1 )

Thus, we can say that 2^{k+2} + 1 is divisible by 3 if 2^{k} + 1 is divisible by 3.

Therefore, we can say that if the given statement is true for n = k, then it is also true for n = k + 2. Hence, by the principle of mathematical induction, the given statement is true ∀ n ∈ O^{+}.