题解 | #等差数列#
等差数列
http://www.nowcoder.com/practice/f792cb014ed0474fb8f53389e7d9c07f
- 等差数列得通项公式 ai = a1+ (i-1)*d;
#include<bits/stdc++.h>
using namespace std;
int main(){
int in;
while(cin>>in){
int a1 = 2, d = 3;
int sum = 0;
for(int i = 1; i <= in;i++){
sum += a1+ (i-1)*d;
}
cout<<sum<<endl;
}
return 0;}
```
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