# Sum of series 1*1*2! + 2*2*3! + ……..+ n*n*(n+1)!

Given n, we need to find sum of 1*1*2! + 2*2*3! + ……..+ n*n*(n+1)!

Examples:

Input: 1
Output: 2

Input: 3
Output: 242

We may assume that overflow does not happen.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A simple solution is to compute terms one by one and add to result.

An efficient solution is based on direct formula 2 + (n*n + n – 2) * (n + 1)!
The working of formula is based on this post.

## C++

 `// CPP program to find sum of the series. ` `#include ` `using` `namespace` `std; ` ` `  `int` `factorial(``int` `n) ` `{ ` `    ``int` `res = 1; ` `    ``for` `(``int` `i = 2; i <= n; i++) ` `        ``res = res * i; ` `    ``return` `res; ` `} ` ` `  `// Function to calculate required series ` `int` `calculateSeries(``int` `n) ` `{ ` `    ``return` `2 + (n * n + n - 2) * factorial(n + 1); ` `} ` ` `  `// Drivers code ` `int` `main() ` `{ ` `    ``int` `n = 3; ` `    ``cout << calculateSeries(n); ` `    ``return` `0; ` `} `

## Java

 `// java program to find sum of the series. ` `import` `java.io.*; ` ` `  `class` `GFG { ` `     `  `    ``static` `int` `factorial(``int` `n) ` `    ``{ ` `        ``int` `res = ``1``; ` `        ``for` `(``int` `i = ``2``; i <= n; i++) ` `            ``res = res * i; ` `        ``return` `res; ` `    ``} ` `     `  `    ``// Function to calculate required series ` `    ``static` `int` `calculateSeries(``int` `n) ` `    ``{ ` `        ``return` `2` `+ (n * n + n - ``2``)  ` `                      ``* factorial(n + ``1``); ` `    ``} ` `     `  `    ``// Drivers code ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{ ` `        ``int` `n = ``3``; ` `        ``System.out.println(calculateSeries(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by anuj_67. `

## Python3

 `# Python program to find sum of ` `# the series. ` `import` `math ` ` `  `def` `factorial(n): ` `    ``res ``=` `1` `    ``i ``=` `2` `    ``for` `i ``in` `(n``+``1``): ` `        ``res ``=` `res ``*` `i ` `    ``return` `res ` `     `  `# Function to calculate required ` `# series ` `def` `calculateSeries(n): ` `    ``return` `(``2` `+` `(n ``*` `n ``+` `n ``-` `2``)  ` `        ``*` `math.factorial(n ``+` `1``)) ` ` `  `# Driver code ` `n ``=` `3` `print``(calculateSeries(n)) ` ` `  `# This code is contributed by  ` `# Prateek bajaj `

## C#

 `// C# program to find sum of the series. ` `using` `System; ` `class` `GFG { ` `     `  `    ``static` `int` `factorial(``int` `n) ` `    ``{ ` `        ``int` `res = 1; ` `        ``for` `(``int` `i = 2; i <= n; i++) ` `            ``res = res * i; ` `        ``return` `res; ` `    ``} ` `     `  `    ``// Function to calculate required series ` `    ``static` `int` `calculateSeries(``int` `n) ` `    ``{ ` `        ``return` `2 + (n * n + n - 2)  ` `                 ``* factorial(n + 1); ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main ()  ` `    ``{ ` `        ``int` `n = 3; ` `        ``Console.WriteLine(calculateSeries(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by anuj_67. `

## PHP

 ` `

Output:

```242
```

My Personal Notes arrow_drop_up Working in Coviam As Backend Developer

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 Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : vt_m, Prateek Bajaj

Article Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.