Question

Write a recursive method:

public static int raise2ToN(int n)

//computes and returns the value of 2n using

//recursion.

Note: 2^3 = 2 * 2^2 .

2^2 = 2 * 2^1

2^1 = 2

What value/values of n will stop the recursion (base case)?

Sample output:

5

2 to 5 is 32

Answer 1

PROGRAM:

import java.util.*;
public class Main
{
   public static int raise2ToN(int n)
{
if(n==0)
{
return 1;
}
else
{
return 2*raise2ToN(n-1);
}
}
   public static void main(String[] args)
   {
       Scanner src=new Scanner(System.in);
       int num=0;
       System.out.print("Enter the value of n: ");
       num=src.nextInt();
       int res=raise2ToN(num);
       System.out.println("2 to "+num+" is "+res);
   }
}
OUTPUT:

This Programs contains a recursive method, which accepts the n value and calculate the 2ⁿ value .

In the method, we are continuously multiplying with 2 and decreasing the value of n until we reach the value of n as 0.

Hope this helps!!!