The sum of the first n positive integers is s u m left paren

The sum of the first n positive integers is:

s u m left parenthesis n right parenthesis equals sum subscript i equals 1 end subscript superscript n i

and can easily be computed using a for-loop:

public int sumToN(int n) {

int sum = 0;

for (int i = 1; i <= n; i++)

{ sum += i; }

return sum; }

It is also possible to recursively compute the sum by recognizing that sum(n) = n + sum(n - 1).

Write a recursive version of sumToN(...).

Solution

SumOfNValues.java

import java.util.Scanner;

public class SumOfNValues {

   public static void main(String[] args) {
      
       //Declaring variables
       int n,sum=0;
      
       //Scanner class object is used to read the values entered by the user
               Scanner sc = new Scanner(System.in);
              
               //Getting the number entered by the user
               System.out.print(\"Enter no of Positive Integers You want to add :\");
               n=sc.nextInt();
              
               //Calling te method by passing the number as argument
               sum=SumOfNos(n);
              
               //Displaying the sum of the numbers till the number entered by the user
               System.out.println(\"Sum of first \"+n+\" numbers is :\"+sum);

   }

   //This method is used to return the sum of numbers recursively.
   private static int SumOfNos(int n) {
if(n>0)
       return n+SumOfNos(n-1);
else
   return 0;
      
   }

}

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Output:

Enter no of Positive Integers You want to add :8
Sum of first 8 numbers is :36

__________Thank You