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

• Last Updated : 18 Mar, 2021

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```

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Simple Solution One by one add elements recursively.
Below is the implementation

## C++

 `// C++ implementation of the approach``#include ``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);``}`

## Java

 `// 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));``    ``}``}`

## Python3

 `# 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`

## C#

 `// 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));``    ``}``}`

## PHP

 ``

## Javascript

 ``
Output:
`8`

Time Complexity: O(n)
Efficient Solution We can solve this problem using direct formula.
Sum can be written as below
∑(n * (n+1))
∑(n*n + n)
= ∑(n*n) + ∑(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++

 `// C++ implementation of the approach``#include ``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);``}`

## Java

 `// 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`

## Python3

 `# 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`

## C#

 `// 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`

## PHP

 ``

## Javascript

 ``
Output:
`8`

Time Complexity: O(1)

My Personal Notes arrow_drop_up