Summing the sum series

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Defined a function that calculates the twice of sum of first N natural numbers as sum(N). Your task is to modify the function to sumX(N, M, K) that calculates sum( K + sum( K + sum( K + …sum(K + N)…))), continuing for M terms. For a given N, M and K calculate the value of sumX(N, M, K).
Note: Since the answer can be very large, print the answer in modulo 10^9 + 7.
Examples:

Input: N = 1, M = 2, K = 3
Output: 552
For M = 2
sum(3 + sum(3 + 1)) = sum(3 + 20) = 552.
Input: N = 3, M =3, K = 2
Output: 1120422
For M = 3
sum(2 + sum(2 + sum(2 + 3))) = sum(2 + sum(2 + 30)) = sum(2 + 1056) = 1120422.

Approach:

Calculate value of sum(N) using the formula N*(N + 1).
Run a loop M times, each time adding K to the previous answer and applying sum(prev_ans + K), modulo 10^9 + 7 each time.
Print the value of sumX(N, M, K) in the end.

Below is the implementation of the above approach:

C++

// C++ program to calculate the  
// terms of summing of sum series
 
#include <iostream>
 
using namespace std;
# define MOD 1000000007
 
// Function to calculate 
// twice of sum of first N natural numbers
long sum(long N){
     
    long val = N * (N+1);
    val = val % MOD;
     
    return val;
}
 
// Function to calculate the 
// terms of summing of sum series
int sumX(int N, int M, int K){
     
    for (int i = 0; i < M; i++) { 
        N = (int)sum(K + N); 
    } 
     
    N = N % MOD; 
    return N;
}
 
// Driver Code
int main()
{
    int N = 1, M = 2, K = 3;
    cout << sumX(N, M, K) << endl; 
    
    return 0;
}
 
// This code is contributed by Rituraj Jain

Java

// Java program to calculate the 
// terms of summing of sum series
 
import java.io.*;
import java.util.*;
import java.lang.*;
 
class GFG {
 
    static int MOD = 1000000007;
 
    // Function to calculate
    // twice of sum of first N natural numbers
    static long sum(long N)
    {
        long val = N * (N + 1);
 
        // taking modulo 10 ^ 9 + 7
        val = val % MOD;
 
        return val;
    }
 
    // Function to calculate the
    // terms of summing of sum series
    static int sumX(int N, int M, int K)
    {
        for (int i = 0; i < M; i++) {
            N = (int)sum(K + N);
        }
        N = N % MOD;
        return N;
    }
 
    // Driver code
    public static void main(String args[])
    {
        int N = 1, M = 2, K = 3;
        System.out.println(sumX(N, M, K));
    }
}

Python3

# Python3 program to calculate the  
# terms of summing of sum series 
   
MOD = 1000000007
   
# Function to calculate 
# twice of sum of first N natural numbers 
def Sum(N): 
      
    val = N * (N + 1) 
   
    # taking modulo 10 ^ 9 + 7 
    val = val % MOD 
   
    return val 
   
# Function to calculate the 
# terms of summing of sum series 
def sumX(N, M, K): 
      
    for i in range(M):
        N = int(Sum(K + N)) 
          
    N = N % MOD 
    return N 
   
if __name__ == "__main__":
      
    N, M, K = 1, 2, 3
    print(sumX(N, M, K)) 
 
# This code is contributed by Rituraj Jain

C#

// C# program to calculate the 
// terms of summing of sum series
 
using System;
class GFG {
 
    static int MOD = 1000000007;
 
    // Function to calculate
    // twice of sum of first N natural numbers
    static long sum(long N)
    {
        long val = N * (N + 1);
 
        // taking modulo 10 ^ 9 + 7
        val = val % MOD;
 
        return val;
    }
 
    // Function to calculate the
    // terms of summing of sum series
    static int sumX(int N, int M, int K)
    {
        for (int i = 0; i < M; i++) {
            N = (int)sum(K + N);
        }
        N = N % MOD;
        return N;
    }
 
    // Driver code
    public static void Main()
    {
        int N = 1, M = 2, K = 3;
        Console.WriteLine(sumX(N, M, K));
    }
}
 
// This code is contributed by anuj_67..

PHP

<?php
// PHP program to calculate the 
// terms of summing of sum series
 
// Function to calculate twice of 
// sum of first N natural numbers
function sum($N)
{
    $MOD = 1000000007;
    $val = $N * ($N + 1);
    $val = $val % $MOD;
     
    return $val;
}
 
// Function to calculate the terms 
// of summing of sum series
function sumX($N, $M, $K)
{
    $MOD = 1000000007;
    for ($i = 0; $i < $M; $i++) 
    { 
        $N = sum($K + $N); 
    } 
     
    $N = $N % $MOD; 
    return $N;
}
 
// Driver Code
$N = 1;
$M = 2;
$K = 3;
echo (sumX($N, $M, $K)); 
     
// This code is contributed 
// by Shivi_Aggarwal
?>

Javascript

<script>
 
// Javascript program to calculate the 
// terms of summing of sum series
 
// Function to calculate twice of 
// sum of first N natural numbers
function sum(N)
{
    let MOD = 1000000007;
    let val = N * (N + 1);
    val = val % MOD;
     
    return val;
}
 
// Function to calculate the terms 
// of summing of sum series
function sumX(N, M, K)
{
    let MOD = 1000000007;
    for (let i = 0; i < M; i++) 
    { 
        N = sum(K + N); 
    } 
     
    N = N % MOD; 
    return N;
}
 
// Driver Code
let N = 1;
let M = 2;
let K = 3;
document.write (sumX(N, M, K)); 
     
// This code is contributed 
// by Sravan
 
</script>

Output:

Time Complexity: O(M)

Auxiliary Space: O(1), since no extra space has been taken.

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