Program for sum of n natural numbers (original) (raw)

Last Updated : 9 Apr, 2026

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

Try It Yourself

Table of Content

• [Naive Approach] Using Loop - O(n) Time and O(1) Space
• [Alternative Approach] Using Recursion -O(n) and O(n) Space
• [Expected Approach] Formula Based Method- O(1) Time and O(1) Space

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

• Initialize sum as 0
• Run a loop from i equals to 1 to n. Inside the loop add i to sum.

For example n = 4
Initially : sum = 0
i = 1, sum = sum + 1 = 1
i = 2, sum = sum + 2 = 3
i = 3, sum = sum + 3 = 7
i = 4, sum = sum + 4 = 11

C++ `

#include 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; }

C

#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; }

Java

import java.io.*;

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));
} 

}

Python

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))

C#

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));
} 

}

JavaScript

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));

`

[Alternative Approach] Using Recursion -O(n) and O(n) Space

In this approach, 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, effectively building the result in a top-down manner.

C++ `

#include 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; }

Java

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)); } }

Python

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

if name == "main": n = 5 print(findSum(n))

C#

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));
}

}

JavaScript

function findSum(n) { // base condition if (n == 1) return 1; return n + findSum(n - 1); } //Driver code let n = 5; console.log(findSum(n));

`

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

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

For example: n = 5
Sum = (5 * (5 + 1)) / 2 = (5 * 6) / 2 = 30 / 2 = 15

**How does this work?

**We can prove this formula using induction.
It is true for n = 1 and n = 2
For n = 1, sum = 1 * (1 + 1)/2 = 1
For n = 4, sum = 4* (4 + 1)/2 = 10
Let it be true for k = n-1.
Sum of k numbers = (k * (k+1))/2
Putting k = n-1, we get
Sum of k numbers = ((n-1) * (n-1+1))/2
= (n - 1) * n / 2
If we add n, we get,
Sum of n numbers = n + (n - 1) * n / 2
= (2n + n2 - n)/2
= n * (n + 1)/2

C++ `

#include 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; }

C

#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; }

Java

import java.io.*; 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)); } }

Python

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))

C#

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));
}

}

JavaScript

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));

`