Find m-th summation of first n natural numbers. - GeeksforGeeks

22 Mar, 2021
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++
Java
C#
PHP
Javascript
// CPP program to find m-th summation#include <bits/stdc++.h>using namespace std; // Function to return mth summationint 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 programint 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.
Python3 

# 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

// 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 summationfunction 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 summationfunction 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
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nitin mittal
vt_m
Rajput-Ji
Mathematical
Recursion
Mathematical
Recursion
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