C++ Program to Find the sum of n natural numbers using Recursion

The numbers 1, 2, 3,…, n are known as natural numbers. This program takes the value of n (entered by user) and prints the sum of first n natural numbers.
For example: If user enters the value of n as 6 then this program would display the sum of first 6 natural numbers:
1+2+3+4+5+6 = 21

In this program we are using recursion to find the sum, we can also solve this problem using loops: C++ program to find the sum of n natural numbers using loop.

Example: Program to calculate and display the sum of n natural numbers using recursion

To understand this program, you should have the knowledge of C++ recursion, if-else statement and functions.

The logic we are using in this program is:
sum(5) = 5+sum(4) = 5+4+sum(3)=5+4+3+sum(2) = 5+4+3+2+sum(1) = 5+4+3+2+1+sum(0) = 15
So the recursive function should look like this: sum(n) = n+sum(n-1)

#include<iostream>
using namespace std;
/* This is function declaration, When you define function
 * after the main then you need to declare it like this.
 * If you define the function before main then no need to
 * declare function.
 */
int sum(int n);
int main(){
   int n;
   cout<<"Enter the value of n(should be positive integer): "; 
   cin>>n;
   /* Here we are checking whether the entered value of n is 
    * natural number or not. If user enters the zero or negative   
    * value then display error message else prints the sum of n    
    * natural numbers.     
    */  
   if(n<=0){
      cout<<"The entered value of n is invalid"; 
   } 
   else{ 
      cout<<"Sum of n natural numbers is:  "<<sum(n); 
   }
   return 0;
}
int sum(int n){ 
   /* We are calling sum function recursively until the value
    * of n is equal to 0.
    */ 
   if(n!= 0) {   
      return n + sum(n-1); 
   } 
   return 0;
}

Output:

Enter the value of n(should be positive integer): 5
Sum of n natural numbers is:  15