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.

step-by-step explanation:

ok one would be money, and the other would be refreshments or something similar. if each bag of popcorn is 5.00 and each drink is half of that that means drinks are 2.50 the proper way to do that would to look and see how much of each you can buy without exceeding a limit of 30.00