# 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

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

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

2^3n = 8^n - 1 = (7+1)^n - 1 .

We expand the right..

2^3n = 7^n + nC1*7^(n-1)*1+ nC2*7^(n-2)+...+nCn-1*7+1 -1

2^3n = 7^n + nC1*7^(n-1)*1+ nC2*7^(n-2)+...+nCn-1*7.

2^n = 7{7^(n-1) + nC1 *7^(n-2) +nC2*7^(n-3)+...nCn-1*}.

The right side is a factor of 7 for all n = 1,2,3, .....

So 7 divides 2^n without remainder.

We'll apply the principle of mathematical induction to prove the equality.

We'll put P(n):(2^3n)-1 | 7

For n = 1, we'll have:

P(1): 2^3 - 1 = 7|7

For n = 1, the principle does hold.

Let P(n) to be true for n = k:

P(k): 2^3k - 1 |7

We'll verify if P(k+1) is true:

P(k+1):2^3(k+1) - 1 = (2^3k)*(2^3) - 1

P(k+1):2^3(k+1) - 1 = (2^3k)*(7+1) - 1

P(k+1):2^3(k+1) - 1 = 7*(2^3k) + 2^3k - 1

But 2^3k - 1 = P(k), that is assumed to be true.

P(k+1): P(k) + 7*(2^3k) | 7 (divisible by 7)

**Since P(k) is true, then P(k+1) is true, hence P(n) is true for any n>=1.**