Problem 6 Sum square difference （数学）

Sum square difference

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.

Answer:

25164150

题解：sum1(n) = n(n + 1)/2.

           sum2(n)=n/6 *(2n + 1)(n + 1).

代码：

#include <bits/stdc++.h>using namespace std;typedef long long ll;int main(){ll a=0;a=100*(100+1)/2;cout<<a*a<<endl;ll b=0;for(int i=1;i<=100;i++){b+=i*i;}cout<<b<<endl;cout<<"ans="<<a*a-b<<endl;return 0;}