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 <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

`// 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 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

`<?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.` `?>` |

## Javascript

`<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

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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