# Find m-th summation of first n natural numbers in C++

In this problem, we are given two integers m and n. Our task is to Find m-th summation of the first n natural numbers.

Problem Description: we will find sum of sum of n natural numbers m times. The sum is given by the formula,

if (m > 1),

sum(n, m) = sum( sum(n, (m-1)), 1 )

if (m = 1)

sum(n, m) = sum(n, 1) = sum of n natural numbers

Let’s take an example to understand the problem,

Input: m = 4, n = 2

Output: 231

Explanation:

sum(2, 4)     = sum ( sum(2, 3), 1 )

= sum ( sum ( sum (2, 2), 1) , 1 )

= sum ( sum ( sum (sum( 2, 1), 1) , 1 ), 1)

= sum ( sum ( sum (3, 1) , 1 ), 1)

= sum ( sum ( 6 , 1 ), 1)
= sum (21, 1)

= 231

## Solution Approach −

A simple solution to the problem is by using two nested loops. The outer one will be for m values and the inner one will be for n values. We will update the value of n and we will calculate the sum of n natural number m times.

A more effective approach would be recursively calling the sum of n natural number m times updating the value with the previous sum each time recursion is made.

Algorithm:

Step 1: update sum value with sum = sum(n, m-1)* (sum(n, m-1)+1) / 2 if the sum value of m is greater than one.

Step 2: If the value of m = 1, return sum = n * (n+1) /2.

Step 3: return sum.

## Example

Live Demo

#include <iostream>
using namespace std;

int calcSumN(int n, int m) {

if (m == 1)
return (n * (n + 1) / 2);
return (calcSumN(n, m-1) * (calcSumN(n, m-1) + 1) / 2);
}

int main() {

int n = 4;
int m = 6;
cout<<m<<"-th summation of first "<<n<<" natural numbers is "<<calcSumN(n, m);
return 0;
}

## Output

6-th summation of first 4 natural numbers is 125230148