题解

include <bits/stdc++.h>
using namespace std;
int main(){
	int n;
	cin>>n;
	long long ans = n*(n+1)*(n+n+1)/6;
	cout<<ans<<endl;
	return 0; 
}

1的平方+2的平方+3的平方+...+n的平方 = n(n+1)(2n+1)/6