以下程序的功能是计算数列
1+1/5+1/8+⋯+1/(2+3n)+⋯+1/(1+302)
的和,请补充完整

include< stdio. h>

mainO
int n:
float sum=0
for n=0 n<=100 ;n++)
sum=sum+
printf( % f" , sum)

参考

//以下程序的功能是计算数列
//1+1/5+1/8+⋯+1/(2+3n)+⋯+1/(1+302)
//的和,请补充完整
#include <stdio.h>

int main() {
    int n;
    float sum = 0;
    for (n = 0; n <= 100; n++)
    sum = sum + 1.0/(2+3*n);
    printf( "%f" , sum);

    return 0;
}

供参考：

#include <stdio.h>
int main()
{
    int  n;
    float sum=0;
    for (n = 0;n <= 100; n++){
        if (n == 0)
            sum += 1.0;
        else
            sum += 1.0/(2 + 3 * n);
    }
    printf("%f", sum);
    return 0;
}