欧拉计划（6）Sum square difference

【题目】

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.

【翻译】

前十个自然数的平方和是：

1² + 2² + ... + 10² = 385

前十个自然数的和的平方是：

(1 + 2 + ... + 10)² = 55² = 3025

所以平方和与和的平方的差是3025 

 385 = 2640.

找出前一百个自然数的平方和与和平方的差。

【思路】直接套公式，简单题。

【代码】

void test6()
{
	int n=100;
	int sum1=(n*n)*((n+1)*(n+1))/4;
	int sum2=(n*(n+1)*((2*n)+1))/6;
	cout<<sum1-sum2<<endl;	
}

【答案】本题答案为25164150 。

来源：CSDN

作者：赵卓不凡

链接：https://blog.csdn.net/sgzqc/article/details/45602297