Program to find sum of first n natural numbers

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 following.

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

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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.

chevron_right


Python

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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.

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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.

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?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
?>

chevron_right



Output :

15

An efficient solution is to use 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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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.

chevron_right


Python

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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.

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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.

chevron_right


php

filter_none

edit
close

play_arrow

link
brightness_4
code

<?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
?>

chevron_right



Output:

15


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

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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)
      return (n/2) * (n+1);
  
   // If n is odd, (n+1) must be even
   else 
      return  ((n + 1) / 2) * n;
}
  
// Driver code
int main()
{
  int n = 5;
  cout << findSum(n);
  return 0;

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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.

chevron_right


Python

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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.

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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.

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?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
?>

chevron_right



Output:

15

This article is contributed by Karik. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



My Personal Notes arrow_drop_up

Improved By : Sam007, Akanksha_Rai