Sum of sum of first n natural numbers

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++

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/* 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;
}

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Java

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// 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

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Python3

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# 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.

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C#

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// 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.

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PHP

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<?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.
?>

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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++

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/* 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;
}

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Java

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// 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

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Python3

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# 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.

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C#

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// 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.

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PHP

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<?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.
?>

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Output:

20

Time complexity : O(1)



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