求和系列的求和

定义了一个函数，该函数将前N个自然数之和的两倍计算为sum(N) 。您的任务是将函数修改为sumX(N，M，K) ，以计算sum(K + sum(K + sum(K +…sum(K + N)…))) ，继续M个项。对于给定的N，M和K，计算sumX(N，M，K)的值。
注意：由于答案可能非常大，请以模数10 ^ 9 + 7打印答案。
例子：

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. 
 

方法：

• 使用公式N *(N + 1)计算sum(N)的值。
• 运行M次循环，每次将K添加到上一个答案，然后应用sum(prev_ans + K) ，每次取模10 ^ 9 + 7。
• 最后打印sumX(N，M，K)的值。

下面是上述方法的实现：

C++

// C++ program to calculate the 
// terms of summing of sum series
 
#include 
 
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..

Javascript

 
// 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
 

输出：

552