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.

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++

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// CPP program to find sum of the series.
#include <bits/stdc++.h>
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;
}

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Java

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// 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.

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Python3

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# 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

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C#

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// 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.

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PHP

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<?php
// PHP program to find sum of the series.
  
function factorial( $n)
{
    $res = 1;
    for ( $i = 2; $i <= $n; $i++)
        $res = $res * $i;
    return $res;
}
  
// Function to calculate required series
function calculateSeries( $n)
{
    return 2 + ($n * $n + $n - 2) * 
                 factorial($n + 1);
}
  
    // Driver code
    $n = 3;
    echo calculateSeries($n);
  
// This code is contributed by anuj_67.
?>

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Output:

242


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Improved By : vt_m, Prateek Bajaj