Sum of the series (1*2) + (2*3) + (3*4) + …… upto n terms

Given a value n, the task is to find sum of the series (1*2) + (2*3) + (3*4) + ……+ n terms

Examples:

Input: n = 2
Output: 8
Explanation:
(1*2) + (2*3)
= 2 + 6
= 8

Input: n = 3
Output: 20
Explanation:
(1*2) + (2*3) + (2*4)
= 2 + 6 + 12
= 20

Simple Solution One by one add elements recursively.

Below is the implementation

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return sum
int sum(int n)
{
    if (n == 1) {
        return 2;
    }
    else {
        return (n * (n + 1) + sum(n - 1));
    }
}
  
// Driver code
int main()
{
  
    int n = 2;
    cout << sum(n);
}

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Java

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// Java implementation of the approach
  
class Solution {
  
    // Function to return a the required result
    static int sum(int n)
    {
        if (n == 1) {
            return 2;
        }
        else {
            return (n * (n + 1) + sum(n - 1));
        }
    }
    // Driver code
    public static void main(String args[])
    {
        int n = 2;
        System.out.println(sum(n));
    }
}

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Python3

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# Python3 implementation of the approach
  
# Function to return sum
def sum(n):
  
    if (n == 1):
        return 2;
    else:
        return (n * (n + 1) + sum(n - 1));
  
# Driver code
  
n = 2;
print(sum(n));
  
# This code is contributed by mits

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C#

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// Csharp implementation of the approach
  
using System;
  
class Solution {
  
    // Function to return a the required result
    static int sum(int n)
    {
        if (n == 1) {
            return 2;
        }
        else {
            return (n * (n + 1) + sum(n - 1));
        }
    }
    // Driver code
    public static void Main()
    {
        int n = 2;
        Console.WrieLine(sum(n));
    }
}

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PHP

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<?php
// PHP implementation of the approach
  
// Function to return sum
function sum($n)
{
    if ($n == 1) 
    {
        return 2;
    }
    else 
    {
        return ($n * ($n + 1) + 
                  sum($n - 1));
    }
}
  
// Driver code
$n = 2;
echo sum($n);
  
// This code is contributed by mits
?>

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Output:

8

Time Complexity: O(n)

Efficient Solution We can solve this problem using direct formula.
Sum can be written as below
&Sum;(n * (n+1))
&Sum;(n*n + n)
= &Sum;(n*n) + &Sum;(n)

We can apply the formulas for sum squares of natural number and sum of natural numbers.

= n(n+1)(2n+1)/6 + n*(n+1)/2
= n * (n + 1) * (n + 2) / 3

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return sum
int sum(int n)
{
    return n * (n + 1) * (n + 2) / 3;
}
  
// Driver code
int main()
{
    int n = 2;
    cout << sum(n);
}

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Java

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// Java implementation of the approach
class GFG
{
      
// Function to return sum
static int sum(int n)
{
    return n * (n + 1) * (n + 2) / 3;
}
  
// Driver code
public static void main(String[] args)
{
    int n = 2;
    System.out.println(sum(n));
}
}
  
// This code is contributed by Code_Mech

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Python3

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# Python3 implementation of the approach. 
  
# Function to return sum 
def Sum(n): 
  
    return n * (n + 1) * (n + 2) // 3
  
# Driver code 
if __name__ == "__main__"
  
    n = 2
    print(Sum(n)) 
  
# This code is contributed 
# by Rituraj Jain

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C#

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// C# implementation of the approach
using System;
      
class GFG
{
       
// Function to return sum
static int sum(int n)
{
    return n * (n + 1) * (n + 2) / 3;
}
   
// Driver code
public static void Main(String[] args)
{
    int n = 2;
    Console.WriteLine(sum(n));
}
}
// This code contributed by Rajput-Ji

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PHP

Output:

8

Time Complexity: O(1)



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