Answer :

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

P(0) = 7^{0} – 2^{0} = 0; is divisible by 5.

P(1) = 7^{1} – 2^{1} = 5; is divisible by 5.

P(2) = 7^{2} – 2^{2} = 45; is divisible by 5.

P(3) = 7^{3} – 2^{3} = 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) = 7^{n} – 2^{n} is divisible by 5 is true for each natural number n.

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