# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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

 ` `

Output:

```5
```

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Improved By : shrikanth13, Code_Mech

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