Consider the following sequence of successive numbers of the 2k -th power: 1, 2²^k , 3²^k , 4²^ k , 5²^k , ... Show that the difference between the numbers in this sequence is odd for all k ∈ N.

The difference between two consecutive numbers in this sequence would be something like:

(2n + 1)^(2*k) - (2n)^(2k)

(or in the other order, but is essentially the same)

where n is an integer number.

Then:

2*n is an even number

2*n + 1 is an odd number.

(2n)^(2k) = 2*(2^(2k-1)*n^(2k)) this will be even for any natural number k.

(2n + 1)^(2k)

Let's suppose that k = 1.

Then we have:

(2n + 1)^(2) = (2n + 1)*(2n + 1) = (2n)^2 + 4n + 1 = 2*(2n^2 + 2n) + 1

If (2n^2 + 2n) is an integer (and it is) we can rename it as N

(2n + 1)^(2) = 2*N + 1

It is still odd.

Now if k = 2, we will have:

(2n + 1)^(2*2) = (2n + 1)^(4)

And using what we found above, we can rewrite this as:

(2*N + 1)^2

And we know that this will be odd.

Then we can already see that:

(2n + 1)^(2k) will be odd for any value of k (where k is a natural number)

Then:

(2n + 1)^(2k) -  (2n)^(2k)

Is the difference between an odd number and an even number, and this will always be an odd number.