Skip to content
Related Articles

Related Articles

Improve Article

Sum of first n natural numbers

  • Difficulty Level : Easy
  • Last Updated : 06 Apr, 2021
Geek Week

Given a positive integer n. The task is to find the sum of the sum of first n natural number.
Examples: 
 

Input : n = 3
Output : 10
Sum of first natural number: 1
Sum of first and second natural number: 1 + 2 = 3
Sum of first, second and third natural number = 1 + 2 + 3 = 6
Sum of sum of first three natural number = 1 + 3 + 6 = 10

Input : n = 2
Output : 4

 

A simple solution is to one by one add triangular numbers. 
 

C++




/* CPP program to find sum
 series 1, 3, 6, 10, 15, 21...
and then find its sum*/
#include <iostream>
using namespace std;
 
// Function to find the sum of series
int seriesSum(int n)
{
    int sum = 0;
    for (int i=1; i<=n; i++)
       sum += i*(i+1)/2;
    return sum;
}
 
// Driver code
int main()
{
    int n = 4;
    cout << seriesSum(n);
    return 0;
}

Java




// Java program to find sum
// series 1, 3, 6, 10, 15, 21...
// and then find its sum*/
import java.io.*;
 
class GFG {
         
    // Function to find the sum of series
    static int seriesSum(int n)
    {
        int sum = 0;
        for (int i = 1; i <= n; i++)
        sum += i * (i + 1) / 2;
        return sum;
    }
 
    // Driver code
    public static void main (String[] args)
    {
        int n = 4;
        System.out.println(seriesSum(n));
         
    }
}
 
// This article is contributed by vt_m

Python3




# Python3 program to find sum
# series 1, 3, 6, 10, 15, 21...
# and then find its sum.
 
# Function to find the sum of series
def seriessum(n):
     
    sum = 0
    for i in range(1, n + 1):
        sum += i * (i + 1) / 2
    return sum
     
# Driver code
n = 4
print(seriessum(n))
 
# This code is Contributed by Azkia Anam.

C#




// C# program to find sum
// series 1, 3, 6, 10, 15, 21...
// and then find its sum*/
using System;
 
class GFG {
 
    // Function to find the sum of series
    static int seriesSum(int n)
    {
        int sum = 0;
         
        for (int i = 1; i <= n; i++)
            sum += i * (i + 1) / 2;
             
        return sum;
    }
 
    // Driver code
    public static void Main()
    {
        int n = 4;
         
        Console.WriteLine(seriesSum(n));
    }
}
 
// This article is contributed by vt_m.

PHP




<?php
// PHP program to find sum
// series 1, 3, 6, 10, 15, 21...
// and then find its sum
 
// Function to find
// the sum of series
function seriesSum($n)
{
    $sum = 0;
    for ($i = 1; $i <= $n; $i++)
        $sum += $i * ($i + 1) / 2;
    return $sum;
}
 
// Driver code
$n = 4;
echo(seriesSum($n));
 
// This code is contributed by Ajit.
?>

Javascript




<script>
 
// javascript program to find sum
// series 1, 3, 6, 10, 15, 21...
// and then find its sum*/
 
    // Function to find the sum of series
    function seriesSum(n) {
        var sum = 0;
        for (i = 1; i <= n; i++)
            sum += i * ((i + 1) / 2);
        return sum;
    }
 
    // Driver code
     
        var n = 4;
        document.write(seriesSum(n));
 
 
// This code contributed by Rajput-Ji
 
</script>

Output: 

20

Time complexity : O(n)
An efficient solution is to use direct formula n(n+1)(n+2)/6
Mathematically, we need to find, Σ ((i * (i + 1))/2), where 1 <= i <= n 
So, lets solve this summation, 
 



Sum = Σ ((i * (i + 1))/2), where 1 <= i <= n
    = (1/2) * Σ (i * (i + 1))
    = (1/2) * Σ (i2 + i)
    = (1/2) * (Σ i2 + Σ i)

We know Σ i2 = n * (n + 1) * (2*n + 1) / 6 and 
Σ i = n * ( n + 1) / 2.
Substituting the value, we get,
Sum = (1/2) * ((n * (n + 1) * (2*n + 1) / 6) + (n * ( n + 1) / 2))  
    = n * (n + 1)/2 [(2n + 1)/6 + 1/2]
    = n * (n + 1) * (n + 2) / 6

Below is the implementation of the above approach: 
 

C++




/* CPP program to find sum
 series 1, 3, 6, 10, 15, 21...
and then find its sum*/
#include <iostream>
using namespace std;
 
// Function to find the sum of series
int seriesSum(int n)
{
    return (n * (n + 1) * (n + 2)) / 6;
}
 
// Driver code
int main()
{
    int n = 4;
    cout << seriesSum(n);
    return 0;
}

Java




// java program to find sum
// series 1, 3, 6, 10, 15, 21...
// and then find its sum
import java.io.*;
 
class GFG
{
    // Function to find the sum of series
    static int seriesSum(int n)
    {
        return (n * (n + 1) * (n + 2)) / 6;
    }
 
   // Driver code
    public static void main (String[] args) {
         
        int n = 4;
        System.out.println( seriesSum(n));
         
    }
}
 
// This article is contributed by vt_m

Python3




# Python 3 program to find sum
# series 1, 3, 6, 10, 15, 21...
# and then find its sum*/
 
# Function to find the sum of series
def seriesSum(n):
 
    return int((n * (n + 1) * (n + 2)) / 6)
 
 
# Driver code
n = 4
print(seriesSum(n))
 
# This code is contributed by Smitha.

C#




// C# program to find sum
// series 1, 3, 6, 10, 15, 21...
// and then find its sum
using System;
 
class GFG {
     
    // Function to find the sum of series
    static int seriesSum(int n)
    {
        return (n * (n + 1) * (n + 2)) / 6;
    }
 
    // Driver code
    public static void Main()
    {
 
        int n = 4;
         
        Console.WriteLine(seriesSum(n));
    }
}
 
// This code is contributed by vt_m.

PHP




<?php
// PHP program to find sum
// series 1, 3, 6, 10, 15, 21...
// and then find its sum
 
// Function to find
// the sum of series
function seriesSum($n)
{
    return ($n * ($n + 1) *
           ($n + 2)) / 6;
}
 
// Driver code
$n = 4;
echo(seriesSum($n));
 
// This code is contributed by Ajit.
?>

Javascript




<script>
// javascript program to find sum
// series 1, 3, 6, 10, 15, 21...
// and then find its sum
 
// Function to find the sum of series
function seriesSum(n)
{
    return (n * (n + 1) * (n + 2)) / 6;
}
 
// Driver code
var n = 4;
document.write( seriesSum(n));
 
// This code is contributed by shikhasingrajput
</script>

Output: 
 

20

Time complexity : O(1)
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :