 ### Find the sum of natural numbers with and without recursion ### Select Articles

Find the sum of natural numbers with and without recursion

Program to find the sum of natural numbers with and without recursion is discussed in this article. A number, N is obtained as input and the sum of first N natural numbers is given as output.

Program to find the sum of natural numbers without using recursion

C

#include

int sum_of_natural_numbers(int n)

{

int sum = 0;

for(int i = 1; i <= n; i++)

{

sum += i;

}

return sum;

}

int main()

{

int n;

printf(“\nEnter the number : “);

scanf(“%d”, &n);

printf(“\nSum of %d Natural Numbers is %d\n “,n,sum_of_natural_numbers(n));

return 0;

}

C++

#include

using namespace std;

int sum_of_natural_numbers(int n)

{

int sum = 0;

for(int i = 1; i <= n; i++)

{

sum += i;

}

return sum;

}

int main()

{

int n;

cout << “\nEnter the number : “;

cin >> n;

cout << “\nSum of ” << n>

cout << endl>

return 0;

}

JAVA

import java.util.*;

public class Main

{

static int sum_of_natural_numbers(int n)

{

int sum = 0;

int i;

for(i = 1; i <= n; i++)

{

sum += i;

}

return sum;

}

public static void main(String args[])

{

int n;

System.out.print(“\nEnter the number : “);

Scanner sc = new Scanner(System.in);

n = sc.nextInt();

System.out.println(“Sum of ” + n + “Natural Numbers is ” + sum_of_natural_numbers(n));

}

}

PYTHON 3

def sum_of_natural_numbers(n):

sum = 0

for i in range(1,n+1):

sum = sum + i

return sum

n = int(input(“Enter a number : “))

print(“Sum of”,n,”Natural Numbers is”,sum_of_natural_numbers(n))

Output

Input- Enter the number:6 Output- Sum of 6 Natural Numbers 21

Time Complexity: O(n)

Program to find the sum of natural numbers using recursion

C

// sum of natural numbers using recursion in c

#include

int sum_of_natural_numbers(int n)

{

if(n == 0)

return 0;

return n + sum_of_natural_numbers(n – 1);

}

int main()

{

int n;

printf(“\nEnter the number : “);

scanf(“%d”, &n);

printf(“\nSum of %d Natural Numbers is %d\n “,n,sum_of_natural_numbers(n));

return 0;

}

C++

// C++ program to find sum of natural numbers using recursion

#include

using namespace std;

int sum_of_natural_numbers(int n)

{

if(n == 0)

return 0;

return n + sum_of_natural_numbers(n – 1);

}

int main()

{

int n;

cout << “\nEnter the number : “;

cin >> n;

cout << “\nSum of ” << n>

cout << endl>

return 0;

}

JAVA

// sum of natural numbers usimg recursion java

import java.util.*;

public class Main

{

static int sum_of_natural_numbers(int n)

{

if(n == 0)

return 0;

return n + sum_of_natural_numbers(n – 1);

}

public static void main(String args[])

{

int n;

System.out.print(“\nEnter the number : “);

Scanner sc = new Scanner(System.in);

n = sc.nextInt();

System.out.println(“Sum of ” + n + “Natural Numbers is ” + sum_of_natural_numbers(n));

}

}

PYTHON 3

# Sum of natural numbers using recursion in python

def sum_of_natural_numbers(n):

if(n == 0):

return 0

return n + sum_of_natural_numbers(n – 1)

n = int(input(“Enter a number : “))

print(“Sum of”,n,”Natural Numbers is”,sum_of_natural_numbers(n))

Output

Input- Enter the number:6 Output- Sum of 6 Natural Numbers 21

Time Complexity: O(n)