Program for sum of n natural numbers

Given a positive integer n, find the sum of the first n natural numbers.

Examples : 

Input: n = 3
Output: 6
Explanation: 1 + 2 + 3 = 6

Input: n = 5
Output: 15
Explanation: 1 + 2 + 3 + 4 + 5 = 15

Sum of Natural Numbers is represented by Triangular Numbers.

[Naive Approach] Loop - O(n) Time and O(1) Space

The idea is to initialize sum to 0 and iterate from 1 to n, adding each number to sum. After the loop ends, return sum.

Consider n = 4

• Initially : sum = 0
• i = 1, sum = sum + 1 = 1
• i = 2, sum = sum + 2 = 3
• i = 3, sum = sum + 3 = 6
• i = 4, sum = sum + 4 = 10

#include <iostream>
using namespace std;

int findSum(int n){
    int sum = 0;
    
    // iterating over all the numbers 
    // between 1 to n
    for (int i = 1; i <= n; i++)
    {
        sum = sum + i;
    }
    return sum;
}
int main()
{
    int n = 5;
    cout << findSum(n);
    return 0;
}

#include <stdio.h>

int findSum(int n){
    int sum = 0;
    
    // iterating over all the numbers 
    // between 1 to n
    for (int i = 1; i <= n; i++)
    {
        sum = sum + i;
    }
    return sum;
}
int main()
{
    int n = 5;
    printf("%d", findSum(n));
    return 0;
}

class GFG{
    
    static int findSum(int n){
        int sum = 0;
        
        // iterating over all the numbers 
        // between 1 to n
        for (int i= 1; i <= n; i++) 
        {
            sum = sum + i;
        }
        return sum;
    }
    public static void main(String args[])
    {
        int n = 5;
        System.out.println(findSum(n));
    } 
}

def findSum(n):
    sum = 0
    i = 1
    
    # Iterating over all the numbers between 1 to n
    while i <= n:
        sum = sum + i
        i = i + 1
    return sum
    
if __name__ == "__main__":
    n = 5
    print(findSum(n))

using System;

class GFG{
    static int findSum(int n)
    {
        int sum = 0;
        
        // iterating over all the numbers 
        // between 1 to n
        for (int i = 1; i <= n; i++) 
        {
            sum = sum + i;
        }
        return sum;
    }
    public static void Main()
    {
        int n = 5;
        Console.Write(findSum(n));
    } 
}

function findSum(n)
{
   let sum = 0;
   
   // iterating over all the numbers 
   // between 1 to n
   for (let i = 1; i <= n; i++) 
   {
     sum = sum + i;
    }
    return sum;
}

// Driver Code
let n = 5;
console.log(findSum(n));

15

[Better Approach] Recursion -O(n) and O(n) Space

We use recursion to find the sum of the first n natural numbers.

• The function calls itself with (n-1) until it reaches the base case of n = 1.
• Each call adds the current value of n to the sum of smaller values.

#include <iostream>
using namespace std;

int findSum (int  n){
    
    // base condition
    if (n == 1 )
     return  1 ;
    return n + findSum(n - 1); 
}
int main() {
    int n = 5 ;
    cout <<  findSum(n); 
    return 0;
}

public class GFG {
    static int findSum(int n) {
        
        // base condition
        if (n == 1)
            return 1;
        return n + findSum(n - 1);
    }
    public static void main(String[] args) {
        int n = 5;
        System.out.println(findSum(n));
    }
}

def findSum(n):
    
    # base condition
    if n == 1:
        return 1
    return n + findSum(n - 1)

if __name__ == "__main__":
    n = 5
    print(findSum(n))

using System;
class GFG
{
    static int findSum(int n)
    {
        // base condition
        if (n == 1)
            return 1;
        return n + findSum(n - 1);
    }

    static void Main()
    {
        int n = 5;
        Console.WriteLine(findSum(n));
    }
}

function findSum(n) {
    
    // base condition
    if (n == 1)
        return 1;
    return n + findSum(n - 1);
}

//Driver code 
let n = 5;
console.log(findSum(n));

15

[Expected Approach] Formula Based Method- O(1) Time and O(1) Space

The sum of the first n natural numbers can be calculated directly using the formula: (n * (n+1)) / 2

We can prove this formula using mathematical induction.

Base Case:

• For n = 1,
• Sum = 1 * (1 + 1) / 2 = 1
• Thus, the formula holds for (n = 1).

Induction Hypothesis:

• Assume the formula is true for n = k, i.e.,
• Sum of first k numbers = k * (k + 1) / 2

Induction Step:

• Putting k = n - 1, we get
• Sum of first n - 1 numbers = (n - 1) * n / 2
• If we add n to this sum, we get,

Sum of first n numbers = n + (n - 1) * n / 2

= (2n + n² - n) / 2

= n * (n + 1) / 2

Hence, the formula is proved.

#include<iostream>
using namespace std;

int findSum(int n)
{
    // Using mathematical formula to compute
    // sum of first n natural numbers
    return n * (n + 1) / 2;
}
int main()
{
   int n = 5;
   cout << findSum(n);
   return 0;
} 

#include<stdio.h>
int findSum(int n)
{
    // Using mathematical formula to compute
    // sum of first n natural numbers
    return n * (n + 1) / 2;
}
int main()
{
    int n = 5;
    printf("%d", findSum(n));
    return 0;
} 

class GFG{
    static int findSum(int n)
    {
        // Using mathematical formula to compute
        // sum of first n natural numbers
        return n * (n + 1) / 2;
    }
    public static void main(String args[])
    {
        int n = 5;
        System.out.println(findSum(n));
    }
}

def findSum(n):
    
    # Using mathematical formula to compute
    # sum of first n natural numbers
    return n * (n + 1) // 2
    
if __name__ == "__main__":
    n = 5
    print(findSum(n))

using System;

class GFG
{
    static int findSum(int n)
    {
        // Using mathematical formula to compute
        // sum of first n natural numbers
        return n * (n + 1) / 2;
    }

    public static void Main()
    {
        int n = 5;
        Console.Write(findSum(n));
    }
}

function findSum(n)
{
    // Using mathematical formula to compute
    // sum of first n natural numbers
    return n * (n + 1) / 2;
}

// Driver Code
var n = 5;
console.log(findSum(n));

15