Project Euler - Problem 6

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.

计算数列和的平方与平方和的差，当n的值很大时，速度会很慢。

def Caculate(n):    sum_sq = sum = 0    for i in range(1, n+1):        sum += i        sum_sq += i ** 2    print sum ** 2 - sum_sq

根据等差数列的求和公式 1 + 2 + 3 + …… + n = n * (n+1) / 2

根据等差数列的平方和公式 1^2 + 2^2 + 3^2 + …… + n^2 = n * (n+1) * (2n+1) / 6

def Cacl(n):    return (n*(n+1)/2)**2 - (n*(n+1)*(2n+1)/6)**2