Why instructor has equated 2^(k) = 1, it should be 2^(k-1) = 1, because first binSearch operates on n, than n/2 and at kth time it shout be n/2^(k-1)
Binary Search time Complexity
Considering there are k steps/iteration. So for the 1st step we say n/2, 2nd step n/4…and so on…This way for the Kth step it is n/2^k.
Here our goal was to find time complexity which is an approximated value. Even if you consider n/2^(k-1), final time complexity would still be O(log n).