题解 | #等差数列#
等差数列
https://www.nowcoder.com/practice/f792cb014ed0474fb8f53389e7d9c07f
#include <iostream>
using namespace std;
int main() {
int n;
while (cin >> n) {
int last_num = 2 + 3*(n - 1);
int ret = 0;
ret = (2 + last_num)*n/2;
cout << ret << endl;
}
}
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