使用递归的系列1 ^ 1 + 2 ^ 2 + 3 ^ 3 +….. + n ^ n的总和

给定一个整数n ，任务是使用递归找到序列1 ¹ + 2 ² + 3 ³ +….. + n ⁿ的总和。
例子：

Input: n = 2 
Output: 5 
1¹ + 2² = 1 + 4 = 5 
Input: n = 3 
Output: 32 
1¹ + 2² + 3³ = 1 + 4 + 27 = 32 
 

方法：从n开始，开始将系列的所有项逐个加法，其中n的值在每个递归调用中递减1 ，直到n = 1的值返回1等于1 ¹ = 1 。
下面是上述方法的实现：

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));
    }
}

Javascript

 
 
// 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.

输出：

5