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 :
 

// 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;
}

// 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.

# Python program to find m-th summation.
def SUM(n, m):
  # base case
  if m == 1:
      return (n * (n + 1) // 2)
  su = SUM(n, m - 1)
  return (su * (su + 1) // 2)

# Driver code
n = 5
m = 3
print("SUM("+str(n)+", "+str(m)+"): "+str(SUM(n, m)))

  

// 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.

<?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.
?>

<script>
// javascript 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);
        
    let sum = SUM(n, m-1);
    return (sum * (sum + 1) / 2);
}

// driver program

    let n = 5;
    let m = 3;
   document.write( "SUM(" + n + ", " + m 
         + "): " + SUM(n, m));

// This code contributed by Rajput-Ji 

</script>

Output: 
 

SUM(5, 3): 7260

Space complexity :- O(M)