Answer :

Let P(n) = 2^{3n} – 1,

For n = 1,

P (1) = 2^{3(1)} -1

= 8-1

= 7

Which is divisible by 7.

Hence P (1) is true.

Consider P(k) to be true.

i.e. 2^{3(k)} -1 is divisible by 7.

⇒ 2^{3(k)} -1 = 7λ for some λ ∈ N

We have to prove that P(k+1) is true i.e. 2^{3(k+1)} -1 is divisible by 7.

Now,

2^{3(k+1)} -1 = 2^{3k} × 2^{3} – 1

= (7λ +1) 2^{3} – 1

= 56λ + 8 – 1

= 7(8λ +1)

Which is divisible by 7.

∴ P(k+1) is true when P(k) is true.

Hence by Mathematical induction P(n) is true for all n ∈ N.

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