If K is a positive integer, then 2 times the 2K power of (- 2) and the 2K + 1 power of (- 2) =?

If K is a positive integer, then 2 times the 2K power of (- 2) and the 2K + 1 power of (- 2) =?

2K power of 2x (- 2) + (2k + 1) power of (- 2)=
=2K power of 2x2 - 2K power of 2x2
=0

If K is a positive integer, then the 2K power of (- 1) and the 2K + 1 power of (- 1) are equal to

2K is even and 2K + 1 is odd
So the original formula = 1 + (- 1) = 0

What is the sum of the (- 2k-1) power of 2 minus the (- 2K + 1) power of 2 plus the (- 2K) power of 2
None of the answers?

The (- 2k-1) power of 2 = 2 ^ (- 2K) / 2
The (- 2K + 1) power of 2 = 2 ^ (- 2K) * 2
The (- 2k-1) power of 2 minus the (- 2K + 1) power of 2 = 2 ^ (- 2K) / 2-2 ^ (- 2K) * 2 = 3 / 2 * 2 ^ (- 2K)
3/2*2^(-2k)+2^(-2k)=5/2*2^(-2k)