What's the rule? Group of 3 | 2 |1?

Official Solution: -(\frac{1}{3})^{-\frac{1}{5}} , -(\frac{1}{2})^{-\frac{1}{4}} , -(\frac{1}{5})^{-\frac{2}{3}} , -(\frac{1}{2})^{-\frac{1}{3}} , -(-(\frac{1}{25})^{-\frac{1}{3}}) . What is the range of the list of numbers above? A. 0 B. \sqrt {25} C. 5 D. 2\sqrt {25} E. 10 The range of a list is the difference between the largest and smallest elements of the list. Hence, to calculate the range, we must first find the smallest and largest numbers of the list given. For that, let's start by rewriting the numbers in easier format: -(\frac{1}{3})^{-\frac{1}{5}} = -(3)^{\frac{1}{5}} = -\sqrt {3} -(\frac{1}{2})^{-\frac{1}{4}} = -(2)^{\frac{1}{4}} = -\sqrt {2} -(\frac{1}{5})^{-\frac{2}{3}} = -(5)^{\frac{2}{3}} = -\sqrt {5^2} = -\sqrt {25} -(\frac{1}{2})^{-\frac{1}{3}} = -(2)^{\frac{1}{3}} = -\sqrt {2} -(-(\frac{1}{25})^{-\frac{1}{3}}) = -(-(25)^{-\frac{1}{3}}) = -(-(\sqrt {25}) = \sqrt {25} Unlike other numbers, the fifth number, \sqrt {25} is positive and thus obviously the largest. Each of -\sqrt {3} , -\sqrt {2} , and -\sqrt {2} are -1.something, while -\sqrt {25} is a little bit more than -3. Thus, -\sqrt {25} is the smallest one. Therefore, the range is \sqrt {25} - (-\sqrt {25}) = 2\sqrt {25} Answer: D

You can just construct a list from the range object:

my_list = list(range(1, 1001))

This is how you do it with generators in python2.x as well. Typically speaking, you probably don't need a list though since you can come by the value of my_list[i] more efficiently (i + 1), and if you just need to iterate over it, you can just fall back on range.

Also note that on python2.x, xrange is still indexable¹. This means that range on python3.x also has the same property²

^{1print xrange(30)[12] works for python2.x}

^{2The analogous statement to 1 in python3.x is print(range(30)[12]) and that works also.}