The positive numbers 1, 2, 3… are known as natural numbers. The program below takes a positive integer from the user and calculates the sum up to the given number.

You can find the sum of natural numbers using loops as well. However, you will learn to solve this problem using recursion here

## Example: Calculate Sum of Natural numbers using Recursion

`#include`
using namespace std;
int add(int n);
int main()
{
int n;
cout << "Enter a positive integer: ";
cin >> n;
cout << "Sum = " << add(n);
return 0;
}
int add(int n)
{
if(n != 0)
return n + add(n - 1);
return 0;
}

**Output**

Enter an positive integer: 10 Sum = 55

In this program, the number entered by the user is passed to the `add()`

function.

Suppose, 10 is entered by the user. Now, 10 is passed to the `add()`

function. This function adds 10 to the addition result of 9 (10 - 1 = 9).

Next time, 9 is added to the addition result of 8 (9 - 1 = 8). This goes on until the number reaches 0, when the function returns 0.

Now, every function is returned to calculate the end result: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55.

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