题解 | #等差数列#
等差数列
http://www.nowcoder.com/practice/f792cb014ed0474fb8f53389e7d9c07f
直接用数学公式
an=a1+(n-1)*d
sn=(a1+an)*n/2
#include<bits/stdc++.h>
using namespace std;
int main()
{
int a1=2,d=3;
int n,an,sum;
while(cin>>n)
{
an=a1+(n-1)*d;
sum=(a1+an)*n/2;
cout<<sum<<endl;
sum=0;
}
}
