Recursion 递归方法，返回从0到int递增1的数字之和

如果递归方法有问题，它应该输出递增1的数字之和，以获得在命令行参数中输入的任何值。如果这有点奇怪，很抱歉

例如，如果10被输入到args中，那么它应该输出55 这是我的代码，所讨论的方法是sumofNumber

public class RecursionMethods {


public static void main(String[] commandlineArguments) {
if (commandlineArguments.length == 0) {
    System.out.println("Please enter a least one commandline!");
   } 
else {
    Integer number = new Integer(0); // initializes the number
    try {
        number = Integer.parseInt(commandlineArguments[0]);
    } catch (NumberFormatException exception) { // NumberFormatException
        System.out.println(exception + " is not a integer.");
        System.exit(1); // ends the program
    }
    String asterisk = asterisks(number); // return address for asterisks
    System.out.println(asterisk);

    String reverse = reverses(number);
    System.out.println(reverse);

    String counting = count(number);
    System.out.println(counting);

    int sums = sumofNumbers(number);
    System.out.println(sums);

 }

}

  public static String asterisks (Integer number) {
  if (number == 0) {
    return "";
} else {
    return "*" + asterisks(number - 1);
  }
}

public static String reverses (Integer number) {
if (number == 0) {
    return "0";
  }
  return number + " " + reverses(number - 1);
 } 

 public static String count (Integer number) {
if (number == 0) {
    return "0";
  }
 return count(number - 1) + " " + number;  
}       
  //here is where the recursion im having trouble with would go 
 public static int sumofNumbers (Integer number) {
   if (number == 0) {
    return 0;  
   }
  return sum(            






}   

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只需将数字添加到numbernumber-1的总和，如果数字为零，则返回：把每一个都写在一张纸上，看看它实际上是如何工作的。制作一张表格，并在每次函数调用时绘制出每个变量的值。@anu非常感谢您，多亏了您，我才算出来！干得好！很高兴你能自己搞定！