Sum of the series 1^1 + 2^2 + 3^3 + ….. + n^n using recursion

Given an integer n, the task is to find the sum of the series 11 + 22 + 33 + ….. + nn using recursion.

Examples:

Input: n = 2
Output: 5
11 + 22 = 1 + 4 = 5

Input: n = 3
Output: 32
11 + 22 + 33 = 1 + 4 + 27 = 32



Approach: Starting from n, start adding all the terms of the series one by one with the value of n getting decremented by 1 in each recursive call until the value of n = 1 for which return 1 as 11 = 1.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
  
// Recursive function to return
// the sum of the given series
ll sum(int n)
{
  
    // 1^1 = 1
    if (n == 1)
        return 1;
    else
  
        // Recursive call
        return ((ll)pow(n, n) + sum(n - 1));
}
  
// Driver code
int main()
{
    int n = 2;
    cout << sum(n);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach
class GFG {
  
    // Recursive function to return
    // the sum of the given series
    static long sum(int n)
    {
  
        // 1^1 = 1
        if (n == 1)
            return 1;
        else
  
            // Recursive call
            return ((long)Math.pow(n, n) + sum(n - 1));
    }
  
    // Driver code
    public static void main(String args[])
    {
        int n = 2;
        System.out.println(sum(n));
    }
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach
  
# Recursive function to return
# the sum of the given series
def sum(n):
    if n == 1:
        return 1
    else:
  
        # Recursive call
        return pow(n, n) + sum(n - 1)
  
# Driver code
n = 2
print(sum(n))
  
# This code is contributed 
# by Shrikant13

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach
using System;
class GFG {
  
    // Recursive function to return
    // the sum of the given series
    static long sum(int n)
    {
        // 1^1 = 1
        if (n == 1)
            return 1;
        else
  
            // Recursive call
            return ((long)Math.Pow(n, n) + sum(n - 1));
    }
  
    // Driver code
    public static void Main()
    {
        int n = 2;
        Console.Write(sum(n));
    }
}

chevron_right


PHP

Output:

5


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : shrikanth13, Code_Mech