Project Euler 6

题目

Sum square difference

The sum of the squares of the first ten natural numbers is, \[1^2 + 2^2 + ... + 10^2 = 385\] The square of the sum of the first ten natural numbers is, \[(1 + 2 + ... + 10)^2 = 55^2 = 3025\] 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.

解决方案

两个公式：

前\(n\)个数的平方和为\(\dfrac{n(n+1)(2n+1)}{6}\).

前\(n\)个数的和为\(\dfrac{n(n+1)}{2}\).

直接代入公式并计算。

代码

n = 100
ans = (n * (n + 1) // 2) ** 2 - n * (n + 1) * (2 * n + 1) // 6
print(ans)