Answer to Question #144312 in Real Analysis for Junaid Hussain

Question #144312
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?
1
Expert's answer
2020-11-16T19:15:55-0500

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.


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