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++

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// 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;
}

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Java

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// 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

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Python3

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# Python3 implementation of the approach
  
# Function to return the sum of
# the first n natural numbers
def summ(n):
    suum = (n * (n + 1)) // 2
    return suum
  
# 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 = summ(n)
  
    return n
  
# Driver code
n = 2
k = 2
  
print(repeatedSum(n, k))
  
# This code is contributed by Mohit Kumar

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\

C#

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// 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

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Output:

6


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