For any natural n

Given; P(n) = 7ⁿ – 2ⁿ is divisible by 5.

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

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

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

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

Let P(k) = 7^k – 2^k is divisible by 5; ⇒ 7^k – 2^k = 5x.

⇒ P(k+1) = 7^k+1 – 2^k+1

= (5 + 2)7^k – 2(2^k)

= 5(7^k) + 2 (7^k – 2^k)

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

⇒ P(k+1) is true when P(k) is true.

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