Answer to Question #144310 in Calculus for Junaid Hussain

Question

Consider the following sequence of successive numbers of the 2^k-th power: 1, 2^2^k, 3^2^k, 4^2^k, 5^2^k, … Show that the difference between the numbers in this sequence is odd for all k ∈ N?

Answer 1

First, we will note that the difference between an even and an odd number is always odd, same for the difference between an odd and an even number.

Secondly, we will remark that a product of two even numbers is even and a product of two odd numbers is odd. Therefore we have that any (positive integer) power of an odd number is odd and any (positive integer) power of an even number is even (as $n^k = n\times n \times n \times ... \times n , k$ times, we apply the former property to this product).

Now, if we take two consecutive elements of this sequence, say $n^{2^k}, (n+1)^{2^k}$ , we have that either n is even (and then n+1 is odd), or n is odd (and then n+1 is even). Then by applying the former property (for all $k\in \mathbb{N}, k$ is a positive integer) we have that either $n^{2^k}$ is even and $(n+1)^{2^k}$ is odd, or $n^{2^k}$ is odd and $(n+1)^{2^k}$ is even. In any of these two cases the difference between these two consecutive terms of sequence is odd by the first mentioned property.