[Project Euler]Problem 6

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.

问题6：
　　前10个自然数的平方的和为385，前十个自然数的和的平方为3025，他们的差为3025-385=2640。 
　　求前一百个自然数的平方的和与和的平方的差。 

方法一：
　　由题意

def f(n):
    return sum(range(1,n+1))**2-sum(i**2 for i in range(1,n+1))

方法二：

　　(1+2+……+n)²=(n²*(n+1)²)/4
　　(1²+2²+……+n²)=(n*(n+1)*(2n+1))/6 
　　(1+2+……+n)²-(1²+2²+……+n²)=(n²*(n+1)²)/4-(n*(n+1)*(2n+1))/6=(n*(n-1)*(n+1)*(3*n+2))/12

def f2(n):
    return ((n+1)*n*(n-1)*(3*n+2))//12