Let f be the function with one argument - positive integer nfunction f(n)  j = 1   for i=1 to n:    j = 2*j    k = j    while k > 0:      k--      print *My approach:let n = 6breakdown the for cycle breakdown the while cycle -> when i == 1, we print two *, when i == 2, we print four *, etc. So I've assumed that it can be $\sum_{i=1}^n 2^i$ what gives me $\Theta(2^n)$.Am I right or did I something wrong?
Time complexity ( $\Theta$ ) of function
algorithm analysis;asymptotics;runtime analysis;loops