Answer :

Given; P(n) = 2^{3n} – 1 is divisible by 7.

P(0) = 2^{0} – 1 = 0; is divisible by 7.

P(1) = 2^{3} – 1 = 7; is divisible by 7.

P(2) = 2^{6} – 1 = 63; is divisible by 7.

P(3) = 2^{9} – 1 = 512; is divisible by 7.

Let P(k) = 2^{3k} – 1 is divisible by 7;

⇒ 2^{3k} – 1 = 7x.

⇒ P(k+1) = 2^{3(k+1)} – 1

= 2^{3}(7x + 1) – 1

= 56x + 7

= 7(8x + 1) ; is divisible by 7.

⇒ P(k+1) is true when P(k) is true.

∴ By Mathematical Induction P(n) = 2^{3n} – 1 is divisible by7, for all natural numbers n.

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