Sum of Natural Numbers using Recursion

Easy Recursion

Find the sum of the first n natural numbers using recursion.

Sample input

Sample output

Solution

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

print(sum_natural(10))

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

console.log(sumNatural(10));

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

    public static void main(String[] args) {
        System.out.println(sumNatural(10));
    }
}

fun sumNatural(n: Int): Int {
    if (n == 0) return 0
    return n + sumNatural(n - 1)
}

fun main() {
    println(sumNatural(10))
}

func sumNatural(_ n: Int) -> Int {
    if n == 0 { return 0 }
    return n + sumNatural(n - 1)
}

print(sumNatural(10))

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

void main() {
  print(sumNatural(10));
}

#include <iostream>
using namespace std;

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

int main() {
    cout << sumNatural(10) << endl;
    return 0;
}

#include <stdio.h>

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

int main() {
    printf("%d\n", sumNatural(10));
    return 0;
}