Related Articles
Find m-th summation of first n natural numbers.
• Difficulty Level : Medium
• Last Updated : 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++

 `// CPP program to find m-th summation``#include ``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

 `// 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#

 `// 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

 ``

## Javascript

 ``

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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up