系列1 * 1的总和！ + 2 * 2！ +…….. + n * n！

给定n，我们需要找到1 * 1的总和！ + 2 * 2！ +…….. + n * n！
例子：

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

我们可以假设不会发生溢出。

一个简单的解决方案是逐项计算项并加到结果中。
一个有效的解决方案是基于直接公式(n + 1)！ – 1
这个公式如何运作?

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 
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.

Javascript

// 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

输出：

23