Program to Find the Sum of Natural Numbers using Recursion

Program to Find the Sum of Natural Numbers using Recursion

  • Write a program to Find the Sum of Natural Numbers using Recursion in C
  • Write a program to Find the Sum of Natural Numbers using Recursion in C++
  • Write a program to Find the Sum of Natural Numbers using Recursion in Python
  • Write a program to Find the Sum of Natural Numbers using Recursion in PHP
  • Write a program to Find the Sum of Natural Numbers using Recursion in Java
  • Write a program to Find the Sum of Natural Numbers using Recursion in Java Script
  • Write a program to Find the Sum of Natural Numbers using Recursion in C#

Explanation:

Logic to Find the Sum of Natural Numbers Using Recursion

To find the sum of the first n natural numbers using recursion, we use the following approach:

  1. Base Case:
    • If the number n is 0, return 0, as the sum of the first 0 natural numbers is 0.
  2. Recursive Case:
    • For any number n, the sum of the first n natural numbers can be found by adding n to the sum of the first n – 1 natural numbers. Thus, the recursive formula is: Sum(n) = n + Sum(n−1)
  3. Stopping Condition:
    • The recursion stops when n becomes 0.

Program to Find the Sum of Natural Numbers using Recursion

#include <stdio.h>

int sum_of_natural(int n) {
    if (n == 0)
        return 0;
    return n + sum_of_natural(n - 1);
}

int main() {
    int n;
    printf("Enter a number: ");
    scanf("%d", &n);

    printf("Sum of first %d natural numbers is %d\n", n, sum_of_natural(n));
    return 0;
}

#include <iostream>
using namespace std;

int sum_of_natural(int n) {
    if (n == 0)
        return 0;
    return n + sum_of_natural(n - 1);
}

int main() {
    int n;
    cout << "Enter a number: ";
    cin >> n;

    cout << "Sum of first " << n << " natural numbers is " << sum_of_natural(n) << endl;
    return 0;
}

def sum_of_natural(n):
    if n == 0:
        return 0
    return n + sum_of_natural(n - 1)

n = int(input("Enter a number: "))
print(f"Sum of first {n} natural numbers is {sum_of_natural(n)}")

<?php
function sum_of_natural($n) {
    if ($n == 0)
        return 0;
    return $n + sum_of_natural($n - 1);
}

echo "Enter a number: ";
$n = intval(trim(fgets(STDIN)));

echo "Sum of first $n natural numbers is " . sum_of_natural($n) . "\n";
?>

import java.util.Scanner;

public class Main {
    public static int sumOfNatural(int n) {
        if (n == 0)
            return 0;
        return n + sumOfNatural(n - 1);
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        System.out.print("Enter a number: ");
        int n = scanner.nextInt();

        System.out.println("Sum of first " + n + " natural numbers is " + sumOfNatural(n));
    }
}

function sumOfNatural(n) {
    if (n === 0)
        return 0;
    return n + sumOfNatural(n - 1);
}

const n = parseInt(prompt("Enter a number:"));
console.log(`Sum of first ${n} natural numbers is ${sumOfNatural(n)}`);

using System;

class Program {
    static int SumOfNatural(int n) {
        if (n == 0)
            return 0;
        return n + SumOfNatural(n - 1);
    }

    static void Main() {
        Console.Write("Enter a number: ");
        int n = int.Parse(Console.ReadLine());

        Console.WriteLine($"Sum of first {n} natural numbers is {SumOfNatural(n)}");
    }
}

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