Sum of series 11! + 22! + ……..+ n*n!

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Given n, we need to find sum of 1*1! + 2*2! + ……..+ n*n!
Examples:

Input: 1
Output: 1
Input: 3
Output: 23
1 * 1! + 2 * 2! + 3 * 3! = 1 + 4 + 18 = 23

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 (n + 1)! – 1
How does this formula work?

We basically need to compute below sum.
?(i * i!) Where i varies from 1 to n
= ?((i + 1 – 1) * i!)
= ?((i+1) * i!) – ?i!
= ?(i + 1)! – ?(i!)
?(i + 1)! = 2! + 3! + … (n+1)! where 1 <= i <= n —–(1)
?(i!) = 1! + 2! + 3! + … (n)! where 1 <= i <= n —–(2)
Subtracting second from first, we get (n+1)! – 1

C++

// 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 factorial(n + 1) - 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 factorial(n + 1) - 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.
 
def factorial(n):
    res = 1
    for i in range(2, n+1):
        res = res * i
    return res
 
# Function to calculate required
# series
def calculateSeries(n):
    return factorial(n + 1) - 1
 
# Drivers code
n = 3
print(calculateSeries(n))
 
# This code is contributed by 
# Ansu Kumari.

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 factorial(n + 1) - 1;
    }
     
    // Driver code
    static public void Main ()
    {
        int n = 3;
        Console.WriteLine( 
                calculateSeries(n));
    }
}
 
// This code is contributed by ajit.

PHP

<?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($n)
{
    return factorial($n + 1) - 1;
}
 
// Driver code
$n = 3;
echo calculateSeries($n);
 
// This code is contributed 
// by akt_mit
?>

Javascript

<script>
// java script program to find
// sum of the series.
 
function factorial(n)
{
    let res = 1;
    for (let i = 2; i <= n; i++)
        res = res * i;
    return res;
}
 
// Function to calculate
// required series
function calculateSeries(n)
{
    return factorial(n + 1) - 1;
}
 
// Driver code
let n = 3;
document.write( calculateSeries(n));
 
// This code is contributed 
// by sravan kumar
</script>

Output:

Time Complexity : O(n).

Auxiliary Space: O(1)

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