Summing the sum series

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:

552

Time Complexity: O(M)

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