题解 | #等差数列#

思路

1. 等差数列前n项和的公式是关键sn=n*a1+n*(n-1)*d/2

Answer

#include<stdio.h>

int main()
{
    int n;
    scanf("%d",&n);
    printf("%d",n*2+n*(n-1)*3/2);

    return 0;
}