I think the answer should be this:

Since the probability of getting two heads in a row and getting two tails in a row is the same, we only need to figure out the probability of the total. The remaining event is:

1 head 1 tail......due to different order, events should be 2

Therefore the probability=0.5(1-2(1/2)^N)=0.5(1-(1/2)^(N-1))

Do you mean Prob(at least 1 observation of 2 consecutive heads appears) ??

Then since the subsets can be:

1. all tails

2. 1 head 1 tail 1 head 1 tail ....

3. at least 2 heads consecutively appear once

So Pr(3) = 1- Pr(1) - Pr(2) = 1- (1/2)^N - (1/2)^N

= 1 - power(1/2,N-1)