Repeated sum of first N natural numbers

Given two integers N and K, the task is to find the sum of first N natural numbers then update N as the previously calculated sum. Repeat these steps K times and finally print the value of N.
Examples:

Input: N = 2, K = 2 
Output: 6 
Operation 1: n = sum(n) = sum(2) = 1 + 2 = 3 
Operation 2: n = sum(n) = sum(3) = 1 + 2 + 3 = 6
Input: N = 3, K = 2 
Output: 21 

Approach: Find the sum of first N natural numbers using the formula (N * (N + 1)) / 2 then update N with the calculated sum. Repeat these steps exactly K times and print the final value of N.
Below is the implementation of the above approach: 

C++

// 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

// 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

# 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#

// 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

Javascript

<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>

Output:

6

Time Complexity: O(k)

Auxiliary Space: O(1)