题解 | #等差数列#
等差数列
https://www.nowcoder.com/practice/f792cb014ed0474fb8f53389e7d9c07f
#include <stdio.h>
int main()
{
int n=0;
while(scanf("%d",&n)!=EOF)
{
int a=2;
if(1==n)
{
printf("%d\n",a);
}
else
{
printf("%d\n",n*a+((n-1)*n*3)/2);
}
}
return 0;
}
