For any natural number n,

According to the question,

P(n) = 7n – 2n is divisible by 5.

So, substituting different values for n, we get,

P(0) = 70 – 20 = 0 Which is divisible by 5.

P(1) = 71 – 21 = 5 Which is divisible by 5.

P(2) = 72 – 22 = 45 Which is divisible by 5.

P(3) = 73 – 23 = 335 Which is divisible by 5.

Let P(k) = 7k – 2k be divisible by 5

So, we get,

⇒ 7k – 2k = 5x.

Now, we also get that,

⇒  P(k+1)= 7k+1 – 2k+1

= (5 + 2)7k – 2(2k)

= 5(7k) + 2 (7k – 2k)

= 5(7k) + 2 (5x) Which is divisible by 5.

⇒ P(k+1) is true when P(k) is true.

Therefore, by Mathematical Induction,

P(n) = 7n – 2n is divisible by 5 is true for each natural number n.