#include<bits/stdc++.h>
const int M=1E9+7;
using namespace std;
int n,t=0;
int main() {
    cin>>n;
    for (int i=1;i<=n;i++) {
        t=(t+(long long)i*i*i)%M;
    }
    cout<<t;
    return 0;
}

1. Dùng vòng lặp: n <= 1e6

#include <bits/stdc++.h>
using namespace std;
#define ll long long
ll n, mod = 1e9 + 7, s = 0;
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin >> n;
    for (ll i = 1; i <= n; ++i)
    {
        s = (s + (i * i * i)) % mod;
    }
    cout << s;
    return 0;
}

2. Dùng toán : n > 1e6

#include <bits/stdc++.h>
using namespace std;
#define ll long long
ll n;
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin >> n;
    cout << (((n % mod) * (n % mod) * ((n + 1) % mod) * ((n + 1) % mod)) / 4) % mod;
    return 0;
}