题解 | #等差数列#
等差数列
https://www.nowcoder.com/practice/f792cb014ed0474fb8f53389e7d9c07f
#include <stdio.h>
int main()
{
int n=0;
int i=0;
int j=2;
int count=0;
scanf("%d",&n);
if(n>=1 && n<=1000)//判段n是否在正确范围
{
for(i=0;i<n;i++)//前n项
{
count+=j;
j+=3;//公差为3
}
}
printf("%d",count);
return 0;
}
