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Sum of the series 1^1 + 2^2 + 3^3 + ….. + n^n using recursion
  • Difficulty Level : Expert
  • Last Updated : 17 Mar, 2021

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:
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++




// 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;
}

Java




// 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));
    }
}

Python3




# 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

C#




// 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));
    }
}

PHP




<?php
// PHP implementation of the approach
 
// Recursive function to return
// the sum of the given series
function sum($n)
{
 
    // 1^1 = 1
    if ($n == 1)
        return 1;
    else
 
        // Recursive call
        return (pow($n, $n) + sum($n - 1));
}
 
// Driver code
$n = 2;
echo(sum($n));
 
// This code is contributed
// by Code_Mech
?>

Javascript




<script>
 
 
// Javascript implementation of the approach
 
// Recursive function to return
// the sum of the given series
function sum(n)
{
 
    // 1^1 = 1
    if (n == 1)
        return 1;
    else
 
        // Recursive call
        return (Math.pow(n, n) + sum(n - 1));
}
 
// Driver code
var n = 2;
document.write(sum(n));
 
// This code is contributed by rutvik_56.
</script>
Output: 
5

 

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