7 divides 2^{3n}-1

Let the given statement be P(*n*), i.e.**,**

**P( n):** 2

^{3n}– 1 is divisible by 7.

It can be observed that P(*n*) is true for *n* = 1 since 2^{3 × 1 }– 1 = 8 – 1 = 7, which is divisible by 7.

Let P(*k*) be true for some positive integer *k*, i.e.,

**P(***k***):** 2^{3k} – 1 is divisible by 7.

∴2^{3k} – 1 = 7*m*; where *m* ∈ N … (1)

We shall now prove that P(*k* + 1) is true whenever P(*k*) is true.

Consider

Therefore, 2^{3n} – 1 is divisible by 7.

Thus, P(*k* + 1) is true whenever P(*k*) is true.

Hence, by the principle of mathematical induction, statement P(*n*) is true for all natural numbers i.e., *n.*

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