Find the sum of Matrix[i][j] whose values are given by abs(i-j)

Given a matrix of size NxN. Rows and Columns are numbered from 0 to N-1. jth column of ith row contains absolute difference between i and j.
In other words, Matrix[i][j] = abs(i-j) where 0 ≤ i, j < N.
you have to find sum of this matrix 
i.e.,

sum = 0
for i=0 to N-1
    for j=0 to N-1
        sum += Matrix[i][j]

For example:

If the size of the matrix is 2
Matrix will be:

0 1
1 0

Sum = 0 + 1 + 1 + 0
        = 2

If the size of the matrix is 3
Matrix will be:

0 1 2
1 0 1
2 1 0

Sum = 0 + 1 + 2 + 1 + 0 + 1 + 2 + 1 + 0
        = 8


Implementation:

#include <stdio.h> unsigned long long int rec(unsigned long long int N) { if(N==0) return 0; return (N*(N-1)+rec(N-1)); } int main() { unsigned long long int t,i,j,N,sum; scanf("%llu",&t); while(t--) { sum=0; scanf("%llu",&N); sum=rec(N); printf("%llu\n",sum); } return 0; }