I think the formula for the no of comparison should be n/2 + 3*n/4 + 7*n/8 + 15*n/16 + ....
=(2-1)n/2 + (4-1)*n/4 + (8-1)*n/8 + 1(6-1)*n/16.....
=(n-n/2)+(n-n/4)+(n-n/8)+......log n times
=nlogn-(n+n/2+n/4.........)
=nlogn-n+1 is the answer for no of comparisons in worst case but in the given question it is asking about the minimum no of comparisons i.e. in best case for every level it will perform n/2 comparisons logn times.So the answer is nlogn/2.
We have to keep in mind that no of comparisons is not equal to asymptotic time complexity