4n – 1 is divisible by 3, for each natural number n.

According to the question,

P(n) = 4n – 1 is divisible by 3.

So, substituting different values for n, we get,

P(0) = 40 – 1 = 0 which is divisible by 3.

P(1) = 41 – 1 = 3 which is divisible by 3.

P(2) = 42 – 1 = 15 which is divisible by 3.

P(3) = 43 – 1 = 63 which is divisible by 3.

Let P(k) = 4k – 1 be divisible by 3,

So, we get,

⇒ 4k – 1 = 3x.

Now, we also get that,

⇒  P(k+1) = 4k+1 – 1

= 4(3x + 1) – 1

= 12x + 3 is divisible by 3.

⇒ P(k+1) is true when P(k) is true

Therefore, by Mathematical Induction,

P(n) = 4n – 1 is divisible by 3 is true for each natural number n.