Find m-th summation of first n natural numbers.

m-th summation of first n natural numbers is defined as following.

If m > 1
  SUM(n, m) = SUM(SUM(n, m - 1), 1)
Else 
  SUM(n, 1) = Sum of first n natural numbers.

We are given m and n, we need to find SUM(n, m).

Examples:

Input  : n = 4, m = 1 
Output : SUM(4, 1) = 10
Explanation : 1 + 2 + 3 + 4 = 10

Input  : n = 3, m = 2 
Output : SUM(3, 2) = 21
Explanation : SUM(3, 2) 
             = SUM(SUM(3, 1), 1) 
             = SUM(6, 1) 
             = 21


Naive Approach : We can solve this problem using two nested loop, where outer loop iterate for m and inner loop iterate for n. After completion of a single outer iteration, we should update n as whole of inner loop got executed and value of n must be changed then. Time complexity should be O(n*m).

for (int i = 1;i <= m;i++)
{
    sum = 0;
    for (int j = 1;j <= n;j++)
        sum += j;
    n = sum;  // update n
}

Efficient Approach :
We can use direct formula for sum of first n numbers to reduce time.
We can also use recursion. In this approach m = 1 will be our base condition and for any intermediate step SUM(n, m), we will call SUM (SUM(n, m-1), 1) and for a single step SUM(n, 1) = n * (n + 1) / 2 will be used. This will reduce our time complexity to O(m).

int SUM (int n, int m)
{
    if (m == 1)
        return (n * (n + 1) / 2);
    int sum = SUM(n, m-1);
    return (sum * (sum + 1) / 2);
}

Below is the implementation of above idea :

C++

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// CPP program to find m-th summation 
#include <bits/stdc++.h>
using namespace std;
  
// Function to return mth summation
int SUM(int n, int m)
{   
    // base case
    if (m == 1)
        return (n * (n + 1) / 2);
          
    int sum = SUM(n, m-1);
    return (sum * (sum + 1) / 2);
}
  
// driver program
int main()
{
    int n = 5;
    int m = 3;
    cout << "SUM(" << n << ", " << m 
         << "): " << SUM(n, m);
    return 0;
}

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Java

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// Java program to find m-th summation.
class GFG {
      
    // Function to return mth summation
    static int SUM(int n, int m) {
          
        // base case
        if (m == 1)
            return (n * (n + 1) / 2);
      
        int sum = SUM(n, m - 1);
          
        return (sum * (sum + 1) / 2);
    }
      
    // Driver code
    public static void main(String[] args) {
          
        int n = 5;
        int m = 3;
          
        System.out.println("SUM(" + n + ", "
                        + m + "): "    + SUM(n, m));
    }
}
  
// This code is contributed by Anant Agarwal.

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Python3

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# Python3 program to find m-th summation 
  
# Function to return mth summation
def SUM(n, m):
  
    # base case
    if (m == 1):
        return (n * (n + 1) / 2)
          
    sum = SUM(n, m-1)
    return int(sum * (sum + 1) / 2)
  
  
# driver program
n = 5
m = 3
print("SUM(", n, ", ", m, "):", SUM(n, m))
  
# This code is contributed by Smitha Dinesh Semwal

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C#

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// C# program to find m-th summation.
using System;
  
class GFG 
{
      
    // Function to return mth summation
    static int SUM(int n, int m) 
    {
          
        // base case
        if (m == 1)
            return (n * (n + 1) / 2);
      
        int sum = SUM(n, m - 1);
          
        return (sum * (sum + 1) / 2);
    }
      
    // Driver Code
    public static void Main() 
    {
          
        int n = 5;
        int m = 3;
          
        Console.Write("SUM(" + n + ", "
                       + m + "): " + SUM(n, m));
    }
}
  
// This code is contributed by Nitin Mittal.

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PHP

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<?php
// PHP program to find m-th summation 
  
// Function to return
// mth summation
function SUM($n, $m)
      
    // base case
    if ($m == 1)
        return ($n * ($n + 1) / 2);
          
    $sum = SUM($n, $m - 1);
    return ($sum * ($sum + 1) / 2);
}
  
    // Driver Code
    $n = 5;
    $m = 3;
    echo "SUM(" , $n , ", " , $m ,
         "): " , SUM($n, $m);
  
// This code is contributed by vt_m.
?>

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Output:

SUM(5, 3): 7260

This article is contributed by Shivam Pradhan (anuj_charm). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : nitin mittal, vt_m



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