#include<stdio.h>
main()
{
int a,d,n,i,s,an
scanf("%d%d%d",&a,&d,&n)
an=0
s=0
for (i=1i<=ni=i++,a=a+d)
{
an=a+an
s=s+an
printf("%d",s)
}
}
#include<stdio.h>int main(){
int n
int begin = 1,end = 100
for(n = beginn <= endn++){
printf("an = %d",10*n-2)
printf("Sn = %d",5*n*n+3*n)
}
return 0
}
你写的这个公式只能求首项为1，公差为1的等差数列的前n项和。对于一般等差数列，这个公式是求不了的。代码如下：123456789#include <stdio.h>void main(){int nprintf("请输入等差数列的项数n: ")scanf("%d",&n)printf("%d",n*(n+1)/2)}
