# Function part 7 (Recursion 2)

**Example 1:**

#include <iostream> using namespace std; int sum(int n) { if (n == 1) return 1; else return n + sum(n - 1); } int main() { cout << sum(5) << endl; return 0; }

This function calculates the sum of the first `n`

natural numbers. The first `n`

natural numbers are the numbers from 1 to `n`

.

The function works by recursively calling itself to calculate the sum of the first `n - 1`

natural numbers. If the input number `n`

is 1, the function simply returns the number 1, because the sum of the first natural number is 1. Otherwise, the function returns the sum of the input number `n`

and the sum of the first `n - 1`

natural numbers.

Output:

15

**Example 2:**

#include <iostream> using namespace std; int sum(int x, int y) { if (x == y) return x; else return y + sum(x, y - 1); } int main() { cout << sum(4, 6) << endl; return 0; }

This is a recursive function `sum`

. It takes two integer parameters, `x`

and `y`

, which represent the range of numbers to sum. Here’s how the function works:

- The base case: If
`x`

is equal to`y`

, it means we have reached the end of the range, so it returns`x`

. This is the terminating condition for the recursion. - The recursive case: If
`x`

is not equal to`y`

, it adds the current value of`y`

to the result of calling the`sum`

function again with the same`x`

and`y-1`

. This effectively breaks down the sum of the range into smaller subproblems by reducing the upper bound (`y`

) by 1 in each recursive call.

n the `main`

function, it calls the `sum`

function with the arguments `4`

and `6`

and then outputs the result to the console using `cout`

. In this case, it will calculate the sum of integers from 4 to 6 (inclusive) and print the result.

When you run this code, it will output the sum of integers from 4 to 6, which is 4 + 5 + 6 = 15. So, the output of this code will be:

15