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.

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)

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.

#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;  }