#include <iostream>
using namespace std;
int main() {
	int a, b, c;
	cin >> a >> b >> c;
	float sum = (1 + a) * a / 2 + b * (b + 1) * (2 * b + 1) / 6;
	for (float i = 1; i <= c; i++)
		sum += 1 / i;
	cout << sum << endl;
	return 0;
}

由公式：

1. 1+2+...+n=(1+n)*n/2
2. 1^2+2^2+...+n^2=n * (n + 1) * (2 * n + 1) / 6