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.

Now 2^(3n)-1

= (2^3)^n-1

= 8^n - 1

= (8 - 1)(8^(n-1) + 8^(n-2) + ... + 8^0)

We have 8-1 =7

Therefore 7 is a factor of all the terms in (8 - 1)(8^(n-1) + 8^(n-2) + ... + 8^0).

Therefore 2^(3n)-1 is divisible by 7

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