I think question is talking about 2-way merge sort.
At the last level, there will be n/2 comparisons.
At the second last level, there will be n/2 list each conataing 2 elements, worst case comparisons (m+n-1) = 3
there are n/4 list comparisons, each needs 3 comparisons.
total comparisons: 3*n/4
at the third last level, there are n/4 lists each having 8 elements, n/8 pairs for for comparison, worst case number of comparisons = 8+8-1 = 15
at 4th last level, each list contains 16 elements, n/8 total lists, and n/16 pairs for comparisons. worst case comparisons = 16+16-1 = 31
total comparisons =
=n/2 + 3*n/4 + 15*n/8 + 31*n/16 + .... (in worst case)
In question, minimum number of comparisons are asked:
at last level = n/2 comparisons
at second last level = n/2 lists, n/4 pairs, each needs 2 comparisons, total comparisons = (n/4)*2 = (n/2)
Aat third last leevl = n/4 lists, n/8 pairs, best comparisons = 4, total comparisons = (n/8)*4 = n/2
such way, logn terms of (n/2) total cost = (nlogn/2).