Question: What is the base case for the following recursion: def f(b, q, n): if n == 0: return 0 else: return b*q**(n-1) + f(b, q,

What is the base case for the following recursion:

def f(b, q, n): if n == 0: return 0 else: return b*q**(n-1) + f(b, q, n-1) print(f(b=2, q=3, n=4))

a.nis 0

b.nis 1

c.nis greater than 1

d.no base case

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