Sum of natural numbers using recursion
Last Updated :
17 Feb, 2023
Given a number n, find sum of first n natural numbers. To calculate the sum, we will use a recursive function recur_sum().
Examples :
Input : 3
Output : 6
Explanation : 1 + 2 + 3 = 6
Input : 5
Output : 15
Explanation : 1 + 2 + 3 + 4 + 5 = 15
Below is code to find the sum of natural numbers up to n using recursion :
C++
#include <iostream>
using namespace std;
int recurSum(int n)
{
if (n <= 1)
return n;
return n + recurSum(n - 1);
}
int main()
{
int n = 5;
cout << recurSum(n);
return 0;
}
|
Java
import java.util.*;
import java.lang.*;
class GFG
{
public static int recurSum(int n)
{
if (n <= 1)
return n;
return n + recurSum(n - 1);
}
public static void main(String args[])
{
int n = 5;
System.out.println(recurSum(n));
}
}
|
Python
def recurSum(n):
if n <= 1:
return n
return n + recurSum(n - 1)
n = 5
print(recurSum(n))
|
C#
using System;
class GFG
{
public static int recurSum(int n)
{
if (n <= 1)
return n;
return n + recurSum(n - 1);
}
public static void Main()
{
int n = 5;
Console.WriteLine(recurSum(n));
}
}
|
PHP
<?php
function recurSum($n)
{
if ($n <= 1)
return $n;
return $n + recurSum($n - 1);
}
$n = 5;
echo(recurSum($n));
?>
|
Javascript
<script>
function recurSum(n)
{
if (n <= 1)
return n;
return n + recurSum(n - 1);
}
let n = 5;
document.write(recurSum(n));
</script>
|
Output :
15
Time complexity : O(n)
Auxiliary space : O(n)
To solve this question , iterative approach is the best approach because it takes constant or O(1) auxiliary space and the time complexity will be same O(n).
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