Project Euler Problem6

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.


#include <bits/stdc++.h>
using namespace std;

int main()
{
  long long int n=100;
  long long sum1 = (n*(n+1))/2;
  long long sum2 = (n*(n+1)*(2*n+1))/6;
  long long int diff = (sum1*sum1) - sum2;
  cout << diff <<endl; 
 
}