**To Prove:**

2^{3n} – 1 is a multiple of 7

Let us prove this question by principle of mathematical induction (PMI) for all natural numbers

2^{3n} – 1 is a multiple of 7

Let P(n): 2^{3n} – 1 which is a multiple of 7

For n = 1 P(n) is true since 2^{3} – 1 = 8 - 1 = 7, which is multiple of 7

Assume P(k) is true for some positive integer k , ie,

= 2^{3k} – 1 = 7m, where m ∈ N …(1)

We will now prove that P(k + 1) is true whenever P( k ) is true

Consider,

[Adding and subtracting 2^{3}]

= 7 x r , where r = 2^{3 }m + 1 is a natural number

Therefore 2^{3n} - 1 is multiple of 7

Therefore, P (k + 1) is true whenever P(k) is true

By the principle of mathematical induction, P(n) is true for all natural numbers ie, N

**Hence proved**