Demonstration. Prove that (2^3n)-1 is divisible by 7.

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We have to prove that (2^3n)-1 is divisible by 7

For n = 1, we have 2^3 - 1 = 8 - 1 = 7 which is divisible by 7.

Now using induction, we need to show that if the relation holds for any n it also holds for n+1.

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We have to prove that (2^3n)-1 is divisible by 7

For n = 1, we have 2^3 - 1 = 8 - 1 = 7 which is divisible by 7.

Now using induction, we need to show that if the relation holds for any n it also holds for n+1.

Let (2^3n) - 1 be divisible by 7

2^[3(n +1)] - 1

=> 2^(3n - 3) - 1

=> 2^3n/2^3 - 1

=> (2^3n - 2^3)/2^3

=> (2^3n - 1 - 7)/2^3

=> [(2^3n) - 1))/2^3] - [7/2^3]

We had assumed (2^3n) - 1 is divisible by 7, so both the terms for the case of n + 1 are divisible by 7.

This shows that if the relation is true for any n it is also true for (n + 1)

This proves that (2^3n)-1 is divisible by 7 for all n>0

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