在这里，你只需要对i从1到9求和，然后在每个循环中输出这个和。
相反，你应该这样做：

int sum = 0;
for (int i = 1; i <= 10; i++)
{
    sum += i;
}
printf("%d", sum);

对于100到500，这样做：

int sum = 0;
for (int i = 100; i <= 500; i++)
{
    sum += i;
}
printf("%d", sum);

将printf置于循环之外

int sum = 0;
for (int i = 1; i <10; i++)
{
  sum = sum + i;
}
printf("%d", sum);

你应该改变它，这样它就可以打印出最终的总和！

int sum = 0;
for (int i = 1; i <10; i++)
{
    sum = sum + i;
}
 printf("%d", sum);

所以你最终程序（从100到500的总和）将是这样的：

int sum = 0;
for (int i = 100; i <500; i++)
{
    sum = sum + i;
}
 printf("%d", sum);

你需要将print语句移出循环。它在每次循环运行时打印每个结果。

一个简单的方法是：在这里它的程序：

public class unoQuini {
    private int sum = 0;
    private static int menor = 1;
    private static int mayor = 5;
    public int getSum() {
      return sum;
    }
    public void setSum(int sum) {
      this.sum = sum;
    }
    public int getMenor() {
      return menor;
    }
    public void setMenor(int menor) {
      this.menor = menor;
    }
    public int getMayor() {
      return mayor;
    }
    public void setMayor(int mayor) {
      this.mayor = mayor;
    }
    public static void main (String Args[]){
      unoQuini unqui = new unoQuini(); 
      int sum2 = 0;
      for (int i = menor; i<= mayor;i++ ){
        sum2 += i;
        unqui.setSum(sum2);
      }
      System.out.println("Suma = "+ unqui.getSum());
    }
}

重要的部分是：

public static void main (String Args[]){
          unoQuini unqui = new unoQuini(); 
          int sum2 = 0;
          for (int i = menor; i<= mayor;i++ ){
            sum2 += i;
            unqui.setSum(sum2);
          }
          System.out.println("Suma = "+ unqui.getSum());
        }

首先，你不包括10（TO数），而且，你在for循环中打印数字

int SumOfNumbers(int from, int to)
{
    int sum;
    if(from > to)
    {
       sum = from;
       from = to;
       to = sum
    }
    sum = 0;
    for(int i = from; i <= to; ++i)
       sum += i;
    return sum;
}

然后你就可以调用你的函数

sum = SumOfNumbers(from, to);
printf("%d\n");

// Print the sum of First n Natural Number

包含<stdio.h>

public void run（）{

int n;
printf("Enter The Natural Number : ");
scanf("%d \n", n);

int sum = 0;
for (size_t i = 0; i <= n; i++)
{
    sum += i;
}

printf("Sum is : %d \n", sum);

return 0;

}