Project Euler #6: Sum square difference - Hacker Rank Program

Problem Statement

_{This problem is a programming version of Problem 6 from projecteuler.net}
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 N natural numbers and the square of the sum.
Input Format
First line contains T that denotes the number of test cases. This is followed by T lines, each containing an integer, N.
Output Format
Print the required answer for each test case.
Constraints 1≤T≤104 1≤N≤104
Sample Input

2
3
10

Sample Output

22
2640

MY SOLUTION:

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {

    long int t;
    scanf("%ld",&t);
    while(t--)
     {
        long int n;
     scanf("%ld",&n);
        printf("%ld\n",(long int)pow(n*(n+1)/2,2)-(n*(n+1)*(2*n+1))/6);
       
    }
    return 0;
}