Answer to Question #144312 in Real Analysis 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

The consecutive difference is odd , not any two terms.

take 2nd and fourth term their difference is even.

Now for consecutive terms, $n^{2^k}$ and $(n+1)^{2^k}$ one of $n, n+1$ is even and the other is odd. Let without loss of generality, $n$ be odd and $n+1$ is then even. Now even *even =even and odd*odd=odd. Hence $n^{2^{k}}$ is odd and $(n+1)^{2^{k}}$ is even. Hence their difference is odd.