题解 | #等差数列#
等差数列
https://www.nowcoder.com/practice/f792cb014ed0474fb8f53389e7d9c07f
#include <iostream>
using namespace std;
int main() {
int a1 = 2, d = 3;
int n;
cin >> n;
int Sn = n*a1 + (n*(n-1)/2)*d;
cout << Sn << endl;
}
// 64 位输出请用 printf("%lld")