Suma repetida de los primeros N números naturales

Dados dos números enteros N y K , la tarea es encontrar la suma de los primeros N números naturales y luego actualizar N como la suma calculada previamente. Repita estos pasos K veces y finalmente imprima el valor de N.
Ejemplos: 
 

Entrada: N = 2, K = 2 
Salida: 6 
Operación 1: n = suma(n) = suma(2) = 1 + 2 = 3 
Operación 2: n = suma(n) = suma(3) = 1 + 2 + 3 = 6
Entrada: N = 3, K = 2 
Salida: 21 
 

Enfoque: Encuentre la suma de los primeros N números naturales usando la fórmula (N * (N + 1)) / 2 y luego actualice N con la suma calculada. Repita estos pasos exactamente K veces e imprima el valor final de N.
A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of the approach
#include <iostream>
using namespace std;
 
// Function to return the sum of
// the first n natural numbers
int sum(int n)
{
    int sum = (n * (n + 1)) / 2;
    return sum;
}
 
// Function to return the repeated sum
int repeatedSum(int n, int k)
{
 
    // Perform the operation exactly k times
    for (int i = 0; i < k; i++) {
 
        // Update n with the sum of
        // first n natural numbers
        n = sum(n);
    }
 
    return n;
}
 
// Driver code
int main()
{
    int n = 2, k = 2;
 
    cout << repeatedSum(n, k);
 
    return 0;
}

// Java implementation of the approach
class GFG
{
     
    // Function to return the sum of
    // the first n natural numbers
    static int sum(int n)
    {
        int sum = (n * (n + 1)) / 2;
        return sum;
    }
     
    // Function to return the repeated sum
    static int repeatedSum(int n, int k)
    {
     
        // Perform the operation exactly k times
        for (int i = 0; i < k; i++)
        {
     
            // Update n with the sum of
            // first n natural numbers
            n = sum(n);
        }
        return n;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int n = 2, k = 2;
     
        System.out.println(repeatedSum(n, k));
     
    }
}
 
// This code is contributed by AnkitRai01

# Python3 implementation of the approach
 
# Function to return the sum of
# the first n natural numbers
def sum(n):
    sum = (n * (n + 1)) // 2
    return sum
 
# Function to return the repeated sum
def repeatedSum(n, k):
 
    # Perform the operation exactly k times
    for i in range(k):
 
        # Update n with the sum of
        # first n natural numbers
        n = sum(n)
 
    return n
 
# Driver code
n = 2
k = 2
 
print(repeatedSum(n, k))
 
# This code is contributed by Mohit Kumar

// C# implementation of the approach
using System;
     
class GFG
{
     
    // Function to return the sum of
    // the first n natural numbers
    static int sum(int n)
    {
        int sum = (n * (n + 1)) / 2;
        return sum;
    }
     
    // Function to return the repeated sum
    static int repeatedSum(int n, int k)
    {
     
        // Perform the operation exactly k times
        for (int i = 0; i < k; i++)
        {
     
            // Update n with the sum of
            // first n natural numbers
            n = sum(n);
        }
        return n;
    }
     
    // Driver code
    public static void Main (String[] args)
    {
        int n = 2, k = 2;
     
        Console.WriteLine(repeatedSum(n, k));
    }
}
 
// This code is contributed by PrinciRaj1992

<script>
// javascript implementation of the approach
 
// Function to return the sum of
// the first n natural numbers
    function sum(n) {
        var sum = (n * (n + 1)) / 2;
        return sum;
    }
 
    // Function to return the repeated sum
    function repeatedSum(n , k) {
 
        // Perform the operation exactly k times
        for (i = 0; i < k; i++) {
 
            // Update n with the sum of
            // first n natural numbers
            n = sum(n);
        }
        return n;
    }
 
    // Driver code
     
        var n = 2, k = 2;
 
        document.write(repeatedSum(n, k));
 
 
// This code contributed by Rajput-Ji
 
</script>

6

Complejidad de tiempo: O(k)

Espacio Auxiliar: O(1)