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

*print*Print*list*Cite

### 2 Answers

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**

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.**