# 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++

`// 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; ` `} ` |

*chevron_right*

*filter_none*

## 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. ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

## 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. ` |

*chevron_right*

*filter_none*

## 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. ` `?> ` |

*chevron_right*

*filter_none*

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.

## Recommended Posts:

- Find if given number is sum of first n natural numbers
- Find sum of N-th group of Natural Numbers
- Program to find sum of first n natural numbers
- Find the average of first N natural numbers
- Find maximum N such that the sum of square of first N natural numbers is not more than X
- Find the good permutation of first N natural numbers
- Find permutation of first N natural numbers that satisfies the given condition
- Find the number of sub arrays in the permutation of first N natural numbers such that their median is M
- Find ways an Integer can be expressed as sum of n-th power of unique natural numbers
- Sum of sum of first n natural numbers
- LCM of First n Natural Numbers
- Natural Numbers
- Sum of first N natural numbers which are divisible by 2 and 7
- Sum of first N natural numbers which are divisible by X or Y
- Sum of squares of first n natural numbers