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Program to find sum of first n natural numbers

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  • Difficulty Level : Basic
  • Last Updated : 27 Sep, 2022
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Given a number n, find the sum of first natural numbers.

Examples : 

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

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

A simple solution is to do the following. 

1) Initialize : sum = 0
2) Run a loop from x = 1 to n and 
   do following in loop.
     sum = sum + x 

C




// C program to find sum of first
// n natural numbers.
#include <stdio.h>
 
// Returns sum of first n natural
// numbers
int findSum(int n)
{
    int sum = 0;
    for (int x = 1; x <= n; x++)
        sum = sum + x;
    return sum;
}
 
// Driver code
int main()
{
    int n = 5;
    printf("%d", findSum(n));
    return 0;
}


C++




// CPP program to find sum of first
// n natural numbers.
#include <iostream>
using namespace std;
 
// Returns sum of first n natural
// numbers
int findSum(int n)
{
    int sum = 0;
    for (int x = 1; x <= n; x++)
        sum = sum + x;
    return sum;
}
 
// Driver code
int main()
{
    int n = 5;
    cout << findSum(n);
    return 0;
}


Java




// JAVA program to find sum of first
// n natural numbers.
import java.io.*;
 
class GFG{
 
    // Returns sum of first n natural
    // numbers
    static int findSum(int n)
    {
        int sum = 0;
        for (int x = 1; x <= n; x++)
            sum = sum + x;
        return sum;
    }
 
    // Driver code
    public static void main(String args[])
    {
        int n = 5;
        System.out.println(findSum(n));
    }
}
 
// This code is contributed by Nikita Tiwari.


Python




# PYTHON program to find sum of first
# n natural numbers.
 
# Returns sum of first n natural
# numbers
def findSum(n) :
    sum = 0
    x = 1
    while x <=n :
        sum = sum + x
        x = x + 1
    return sum
 
 
# Driver code
 
n = 5
print findSum(n)
 
# This code is contributed by Nikita Tiwari.


C#




// C# program to find sum of first
// n natural numbers.
using System;
 
class GFG{
 
    // Returns sum of first n natural
    // numbers
    static int findSum(int n)
    {
        int sum = 0;
        for (int x = 1; x <= n; x++)
            sum = sum + x;
        return sum;
    }
 
    // Driver code
    public static void Main()
    {
        int n = 5;
        Console.Write(findSum(n));
    }
}
 
// This code is contributed by vt_m.


PHP




<?php
// PHP program to find sum of first
// n natural numbers.
 
// Returns sum of first n natural
// numbers
function findSum($n)
{
$sum = 0;
for ($x = 1; $x <= $n; $x++)
    $sum = $sum + $x;
return $sum;
}
 
// Driver code
$n = 5;
echo findSum($n);
 
// This code is contributed by Sam007
?>


Javascript




<script>
// Javascript program to find sum of first
// n natural numbers.
 
// Returns sum of first n natural
// numbers
function findSum(n)
{
   let sum = 0;
   for (let x = 1; x <= n; x++)
     sum = sum + x;
   return sum;
}
 
// Driver code
let n = 5;
document.write(findSum(n));
 
// This code is contributed by rishavmahato348.
</script>


Output

15

Time Complexity: O(n)

Auxiliary Space: O(1)

An efficient solution is to use the below formula.

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 = 2, sum = 2 * (2 + 1)/2 = 3

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




// Efficient C program to find
// sum of first n natural numbers.
#include<stdio.h>
 
// Returns sum of first n natural
// numbers
int findSum(int n)
{
   return n * (n + 1) / 2;
}
 
// Driver code
int main()
{
  int n = 5;
  printf("%d", findSum(n));
  return 0;
}


C++




// Efficient CPP program to find sum of first
// n natural numbers.
#include<iostream>
using namespace std;
 
// Returns sum of first n natural
// numbers
int findSum(int n)
{
   return n * (n + 1) / 2;
}
 
// Driver code
int main()
{
  int n = 5;
  cout << findSum(n);
  return 0;
}


Java




// Efficient JAVA program to find sum
// of first n natural numbers.
import java.io.*;
 
class GFG{
     
    // Returns sum of first n natural
    // numbers
    static int findSum(int n)
    {
        return n * (n + 1) / 2;
    }
 
    // Driver code
    public static void main(String args[])
    {
        int n = 5;
        System.out.println(findSum(n));
    }
}
 
// This code is contributed by Nikita Tiwari.


Python




# Efficient CPP program to find sum
# of first n natural numbers.
 
# Returns sum of first n natural
# numbers
def findSum(n) :
    return n * (n + 1) / 2
     
# Driver code
n = 5
print findSum(n)
 
# This code is contributed by Nikita Tiwari.


C#




// Efficient C# program to find sum
// of first n natural numbers.
using System;
 
class GFG{
     
    // Returns sum of first n natural
    // numbers
    static int findSum(int n)
    {
        return n * (n + 1) / 2;
    }
 
    // Driver code
    public static void Main()
    {
        int n = 5;
        Console.Write(findSum(n));
    }
}
 
// This code is contributed by vt_m.


php




<?php
// Efficient PHP program to find sum
// of first n natural numbers.
 
// Returns sum of first n natural
// numbers
function findSum($n)
{
    return ($n * ($n + 1) / 2);
}
 
// Driver code
$n = 5;
echo findSum($n);
 
// This code is contributed by Sam007
?>


Javascript




<script>
// javascript Program to find the average
// of sum of first n natural numbers
 
// Return the average of sum
// of first n even numbers
function findSum(n)
{
    return n * (n + 1) / 2;
}
var n = 5;
document.write(findSum(n));
 
// This code is contributed by sravan kumar
</script>


Output

15

Time Complexity: O(1)

Auxiliary Space: O(1)

The above program causes overflow, even if the result is not beyond the integer limit. We can avoid overflow up to some extent by dividing first.

C




// Efficient C program to find
// sum of first n natural numbers
// that avoids overflow if result
// is going to be within limits.
#include<stdio.h>
  
// Returns sum of first n natural
// numbers
int findSum(int n)
{
   if (n % 2 == 0)
      
      // Here multiplying by 1LL help to
      // perform calculations in long long,
      // so that answer should not be overflowed
      return (n / 2) * 1LL * (n + 1);
  
   // If n is odd, (n+1) must be even
   else
      
      // Here multiplying by 1LL help to
      // perform calculations in long long,
      // so that answer should not be overflowed
      return  ((n + 1) / 2) * 1LL * n;
}
  
// Driver code
int main()
{
  int n = 5;
  printf("%d", findSum(n));
  return 0;
}


C++




// Efficient CPP program to find sum of first
// n natural numbers that avoids overflow if
// result is going to be within limits.
#include<iostream>
using namespace std;
  
// Returns sum of first n natural
// numbers
int findSum(int n)
{
   if (n % 2 == 0)
       
      // Here multiplying by 1LL help to
      // perform calculations in long long,
      // so that answer should not be overflowed
      return (n / 2) * 1LL * (n + 1);
  
   // If n is odd, (n+1) must be even
   else
      
      // Here multiplying by 1LL help to
      // perform calculations in long long,
      // so that answer should not be overflowed
      return  ((n + 1) / 2) * 1LL * n;
}
  
// Driver code
int main()
{
  int n = 5;
  cout << findSum(n);
  return 0;
}


Java




// Efficient JAVA program to find sum of first
// n natural numbers that avoids overflow if
// result is going to be within limits.
import java.io.*;
 
class GFG{
 
    // Returns sum of first n natural
    // numbers
    static int findSum(int n)
    {
        if (n % 2 == 0)
            return (n / 2) * (n + 1);
 
        // If n is odd, (n+1) must be even
        else
            return ((n + 1) / 2) * n;
    }
 
    // Driver code
    public static void main(String args[])
    {
        int n = 5;
        System.out.println(findSum(n));
    }
}
 
//This code is contributed by Nikita Tiwari.


Python




# Efficient Python program to find the sum 
# of first n natural numbers that avoid
# overflow if the result is going to be
# within limits.
 
# Returns sum of first n natural
# numbers
def findSum(n) :
    if (n % 2 == 0) :
        return (n / 2) * (n + 1)
  
   # If n is odd, (n+1) must be even
    else :
       return  ((n + 1) / 2) * n
        
# Driver code
n = 5
print findSum(n)
 
# This code is contributed by Nikita Tiwari.


C#




// Efficient C# program to find the sum of first
// n natural numbers that avoid overflow if
// result is going to be within limits.
using System;
 
class GFG{
 
    // Returns sum of first n natural
    // numbers
    static int findSum(int n)
    {
        if (n % 2 == 0)
            return (n / 2) * (n + 1);
 
        // If n is odd, (n+1) must be even
        else
            return ((n + 1) / 2) * n;
    }
 
    // Driver code
    public static void Main()
    {
        int n = 5;
        Console.Write(findSum(n));
    }
}
 
// This code is contributed by vt_m.


PHP




<?php
// Efficient php program to find sum of first
// n natural numbers that avoids overflow if
// result is going to be within limits.
 
// Returns sum of first n natural
// numbers
function findSum($n)
{
    if ($n % 2 == 0)
        return ($n / 2) *
               ($n + 1);
     
    // If n is odd, (n+1) must be even
    else
        return (($n + 1) / 2) * $n;
}
 
// Driver code
$n = 5;
echo findSum($n);
 
// This code is contributed by Sam007
?>


Javascript




<script>
//efficient approach using  javascript to find the average
// of sum of first n natural numbers
 
// Return the average of sum
// of first n even numbers
function findSum(n)
{
    if (n % 2 == 0)
        return (n / 2) * (n + 1)
 
// If n is odd, (n+1) must be even
    else
    return ((n + 1) / 2) * n
         
}
var n = 5;
document.write(findSum(n));
 
// This code is contributed by sravan kumar
</script>


Output

15

Time Complexity: O(1)

Auxiliary Space: O(1)

This article is contributed by Tapesh (tapeshdua420). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks. Please write comments if you find anything incorrect, or if you want to share more information about the topic discussed above.


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