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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