这是打印一系列数字的有效方法（它可以变得更好）吗？

Question

使用高斯方法，可以快速有效地打印一系列数字。 ：）。

使用循环涉及的答案对大数字不利（例如：打印从1到1 ^ 19的数字之和）。

假定用户不会输入错误的输入。

Gauss方法取决于以聪明的方式加数字，您不必遍历脑海中所有的数字，即公式（n + 1）（n / 2），其中n表示您正在经历的数字，从1-> 10，n = 10。

def sum_of_numbers(start_number, end_number):

    # n self explanatory
    n = abs(end_number - start_number) + 1

    # n / 2 represents the number of pairs, end_number+start_number 
    represents the total for each pair.
    return (end_number + start_number) * (n / 2)

def sum_of_odds(start_number, end_number):
    """ start_number < end_number, start_number and end_number are odd """
    # When adding odd numbers (1 + 3 + 5 + 7 + ....... + 2n-1)
    # you will get (1, 4, 9, 16,.......,n**2), the total can be represented as 
    # n**2, in order to find the total of the first, lets say 100 numbers, 
    # we use 100/2**2(half even and half odd) which is 5000, it works!

    # To make this a-bit more general, lets say we want to find the odd 
    # numbers from 5 to 11, its like adding  the whole thing which is 6**2 and 
    # the we subtract the first 2 odd numbers ( 6**2 - 1 - 3)

    start_number_position = (start_number + 1) / 2
    end_number_position = (end_number + 1) / 2

    return (end_number_position)**2 - (start_number_position - 1)**2

def sum_of_evens(start_number, end_number):
   """start_number < end_number, start_number and end_number are even"""

    # To find even numbers total between 2 numbers, (e.g : 12->20)
    # We add the odd numbers before each even number (11+13+15+17+19) then we 
    # add 1 for each number (1*5)
    return sum_of_odds(start_number-1, end_number-1) \
    + (end_number - start_number)/2 + 1

您怎么看？我们可以做得更好吗？