Problem 6

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本文提供了一种计算方法，用于找出前一百个自然数的平方和与这些数之和的平方之间的差值。通过两个独立的函数实现，即求平方和与求和的平方，最后计算两者之差。

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问题描述：
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.

	//求平方和 
	public long squares_sum(int number){
		long result = 0;

		for(int i=1; i<=number; i++){
			result += i*i;
		}
		return result;
	}
	//求和的平方
	public long sum_squares(int number){
		long result = (number*(number+1))/2;
		return result*result;
	}

	public long count(int number){
		return sum_squares(number)-squares_sum(number);
	}