Question

Java, how would i do this

public static void main(String[] args) { int n = 3; int result; result = factorial(n); + public static int factorial(int n) public static int factorial(int n) n = 3 { returns 3* 2 * 1 { if (n == 1) return n; else return (n * factorial(n-1)); if (n == 1) return n; else return (3 * factorial(3-1)); ܢܟ } public static int factorial(int n) n = 2 public static int factorial(int n) returns 2* 1 { if (n == 1) return n; else return (n * factorial(n-1)); if (n == 1) return n; else return (2 * factorial(2-1)); } } n = 1 public static int factorial(int n) { public static int factorial(int n) { if (n == 1) return n; else return (n * factorial(n-1) ); returns 1 if (n == 1) return 1; Do one (or more) of the exercises that produces a program. Exercise 6.6.2: The sum of a series of consecutive numbers from 1 to n can be defined recursively as: sum(1) = 1; sum(n) = n + sum(n-1); Write a recursive Java method that accepts n as an argument and calculates the sum of the numbers from 1 to n. [Hint: modify the sample program (above) so that it returns (n+sum(n-1)]

Answer 1

The answer to this question is as follows:

The code is as follows:

import java.util.*;
public class MyClass {
public static int sum(int n)
{
if(n==1)
return 1;
else
return n+sum(n-1);
}
public static void main(String args[]) {
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
System.out.println(sum(n));
}
}

The input and output are provided in the screenshot below:

1 import java.util.*; 2- public class MyClass { 3 public static int sum(int n) { if(n==1) return 1; else return n+sum(n-1); } 5 6 7 8 9 10 11 12 13 14 15 public static void main(String args[]) { Scanner sc=new Scanner(System.in); int n=sc.nextInt(); System.out.println(sum(n)); } } Execute Mode, Version, Inputs & Arguments JDK 11.0.4 Stdin Inputs Interactive Command Line Arguments 10 Execute ... ca Result CPU Time: 0.20 sec(s), Memory: 34656 kilobyte(s) compiled and executed in 0.804 sec(s) 55