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  • Problem 6

    Problem 6

    # Problem_6.py
    """
    The sum of the squares of the first ten natural numbers is,
    12 + 22 + ... + 102 = 385
    The square of the sum of the first ten natural numbers is,
    (1 + 2 + ... + 10)2 = 552 = 3025
    Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640.
    Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum
    找出100一下的所有自然数的平方和与这些自然数的和的平方之间的差数.
    """
    sum_of_squares = sum([i*i for i in range(1, 101)])
    square_of_sum = pow(sum([i for i in range(1, 101)]), 2)
    
    print(square_of_sum, '-', sum_of_squares, '=', square_of_sum-sum_of_squares)
    Resistance is Futile!
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  • 原文地址:https://www.cnblogs.com/noonjuan/p/10957316.html
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