题解 | #等差数列#
等差数列
https://www.nowcoder.com/practice/f792cb014ed0474fb8f53389e7d9c07f
#include <bits/stdc++.h>
using namespace std;
#define a1 2
#define d 3
int main(){
int n = 0;
while(cin >> n){
int res = 0;
int an = a1 + (n - 1) * d;
res = (a1 + an) * n / 2;
cout << res << endl;
}
return 0;
}
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