Compute the nth Fibonacci number recursively, where fib(0) = 0, fib(1) = 1, and every later term is the sum
of the two before it.
This needs two base cases instead of one, and – new in this exercise – a recursive case that calls itself
twice, not once, each on a smaller input. Try tracing fibonacci(4) by hand, the way the lesson traced
factorial(5): you’ll notice the call tree branches into two calls at every level, and that fibonacci(2) in
particular ends up computed more than once along the way – worth noticing now, even though fixing that
inefficiency is out of scope for this exercise.