Find sum of the series 1! – 2! + 3! – 4! + 5! . . . till Nth term

Given a positive integer N, the task is to find the sum of series 1! – 2! + 3! – 4! + 5!… till Nth term.

Examples:

Input: N = 6
Output: -619
Explanation: The sum of the series upto 5th term can be calculated as
1! – 2! + 3! – 4! + 5! -6! = 1 -2 +6 -24 +120 -720 = -619

Input: N = 5
Output: 101

Native Approach: The simplest way to solve this problem is to find the factorial of all numbers in the range [1, N] and calculate their sum with their respective positive and negative sign, i.e., odd position will be (+)ve and even position will be negative.

Below is the implementation of the above approach:

C++

// C++ program of the above approach
#include <iostream>
using namespace std;

// Function to find factorial
// of a given number
int factorial(int N)
{
    if (N == 1) {
        return 1;
    }

    // Recursive Call
    return N * factorial(N - 1);
}

// Function to find the sum of
// the series 1! - 2! + 3! - 4!
// + 5!... till the Nth term
int SeriesSum(int N)
{
    // Stores Required Sum
    int Sum = 0;

    // Loop to calculate factorial
    // and sum them with their
    // respective sign
    for (int i = 1; i <= N; i++) {

        // Factorial of cur integer
        int fact = factorial(i);

        // Stores the sign
        int sign = fact;

        // If position is even sign
        // must be negative
        if (i % 2 == 0) {
            sign = sign * -1;
        }

        // Adding value in sum
        Sum += sign;
    }

    // Return Answer
    return Sum;
}

// Driver Code
int main()
{
    int N = 6;
    cout << SeriesSum(N);
    return 0;
}

Java

// Java implementation for the above approach
import java.util.*;

class GFG{

// Function to find factorial
// of a given number
static int factorial(int N)
{
    if (N == 1)
    {
        return 1;
    }

    // Recursive Call
    return N * factorial(N - 1);
}

// Function to find the sum of
// the series 1! - 2! + 3! - 4!
// + 5!... till the Nth term
static int SeriesSum(int N)
{

    // Stores Required Sum
    int Sum = 0;

    // Loop to calculate factorial
    // and sum them with their
    // respective sign
    for(int i = 1; i <= N; i++)
    {

        // Factorial of cur integer
        int fact = factorial(i);

        // Stores the sign
        int sign = fact;

        // If position is even sign
        // must be negative
        if (i % 2 == 0)
        {
            sign = sign * -1;
        }

        // Adding value in sum
        Sum += sign;
    }

    // Return Answer
    return Sum;
}

// Driver Code
public static void main(String args[])
{
    int N = 6;

    System.out.print(SeriesSum(N));
}
}

// This code is contributed by sanjoy_62

Python

# Python program of the above approach

# Function to find factorial
# of a given number
def factorial(N):

    if (N == 1):
        return 1

    # Recursive Call
    return N * factorial(N - 1)

# Function to find the sum of
# the series 1! - 2! + 3! - 4!
# + 5!... till the Nth term
def SeriesSum(N):

    # Stores Required Sum
    Sum = 0

    # Loop to calculate factorial
    # and sum them with their
    # respective sign
    for i in range(1, N + 1):

        # Factorial of cur integer
        fact = factorial(i);

        # Stores the sign
        sign = fact;

        # If position is even sign
        # must be negative
        if (i % 2 == 0):
            sign = sign * -1

        # Adding value in sum
        Sum += sign

    # Return Answer
    return Sum

# Driver Code

N = 6
print(SeriesSum(N))

# This code is contributed by Samim Hossain Mondal.

C#

// C# implementation for the above approach
using System;

class GFG{

// Function to find factorial
// of a given number
static int factorial(int N)
{
    if (N == 1)
    {
        return 1;
    }

    // Recursive Call
    return N * factorial(N - 1);
}

// Function to find the sum of
// the series 1! - 2! + 3! - 4!
// + 5!... till the Nth term
static int SeriesSum(int N)
{

    // Stores Required Sum
    int Sum = 0;

    // Loop to calculate factorial
    // and sum them with their
    // respective sign
    for(int i = 1; i <= N; i++)
    {

        // Factorial of cur integer
        int fact = factorial(i);

        // Stores the sign
        int sign = fact;

        // If position is even sign
        // must be negative
        if (i % 2 == 0)
        {
            sign = sign * -1;
        }

        // Adding value in sum
        Sum += sign;
    }

    // Return Answer
    return Sum;
}

// Driver Code
public static void Main()
{
    int N = 6;

    Console.Write(SeriesSum(N));
}
}

// This code is contributed by Samim Hossain Mondal.

Javascript

<script>
       // JavaScript code for the above approach


       // Function to find factorial
       // of a given number
       function factorial(N) {
           if (N == 1) {
               return 1;
           }

           // Recursive Call
           return N * factorial(N - 1);
       }

       // Function to find the sum of
       // the series 1! - 2! + 3! - 4!
       // + 5!... till the Nth term
       function SeriesSum(N) {
           // Stores Required Sum
           let Sum = 0;

           // Loop to calculate factorial
           // and sum them with their
           // respective sign
           for (let i = 1; i <= N; i++) {

               // Factorial of cur integer
               let fact = factorial(i);

               // Stores the sign
               let sign = fact;

               // If position is even sign
               // must be negative
               if (i % 2 == 0) {
                   sign = sign * -1;
               }

               // Adding value in sum
               Sum += sign;
           }

           // Return Answer
           return Sum;
       }

       // Driver Code

       let N = 6;
       document.write(SeriesSum(N));

 // This code is contributed by Potta Lokesh
   </script>

Output

-619

Time Complexity: O(N²)  
Auxiliary Space: O(N) for call stack because using recursion  

Efficient Approach: The above solution can be optimized by maintaining the value of the factorial of the previous number and calculating the factorial of the current number using that value and calculating their sum with their respective positive and negative sign.

Below is the implementation of the above approach: 

C++

// C++ program of the above approach
#include <iostream>
using namespace std;

// Function to find the sum of
// the series 1! - 2! + 3! - 4!
// + 5!... till the Nth term
int SeriesSum(int N)
{
    // Stores the required sum
    // and factorial value
    int Sum = 0, fact = 1;

    // Loop to calculate factorial
    // and sum them with their
    // respective sign
    for (int i = 1; i <= N; i++) {

        // Calculate factorial
        fact = fact * i;

        // Stores the sign
        int sign = fact;

        // If position is even sign
        // must be negative
        if (i % 2 == 0) {
            sign = sign * -1;
        }

        // Adding value in sum
        Sum += sign;
    }

    // Return Answer
    return Sum;
}

// Driver Code
int main()
{
    int N = 6;
    cout << SeriesSum(N);
    return 0;
}

Java

// Java program of the above approach

import java.util.*;

class GFG{

// Function to find the sum of
// the series 1! - 2! + 3! - 4!
// + 5!... till the Nth term
static int SeriesSum(int N)
{
    // Stores the required sum
    // and factorial value
    int Sum = 0, fact = 1;

    // Loop to calculate factorial
    // and sum them with their
    // respective sign
    for (int i = 1; i <= N; i++) {

        // Calculate factorial
        fact = fact * i;

        // Stores the sign
        int sign = fact;

        // If position is even sign
        // must be negative
        if (i % 2 == 0) {
            sign = sign * -1;
        }

        // Adding value in sum
        Sum += sign;
    }

    // Return Answer
    return Sum;
}

// Driver Code
public static void main(String[] args)
{
    int N = 6;
    System.out.print(SeriesSum(N));
}
}

// This code is contributed by 29AjayKumar

Python3

# Python 3 program of the above approach

# Function to find the sum of
# the series 1! - 2! + 3! - 4!
# + 5!... till the Nth term
def SeriesSum(N):

    # Stores the required sum
    # and factorial value
    Sum = 0
    fact = 1

    # Loop to calculate factorial
    # and sum them with their
    # respective sign
    for i in range(1, N + 1):

        # Calculate factorial
        fact = fact * i

        # Stores the sign
        sign = fact

        # If position is even sign
        # must be negative
        if (i % 2 == 0):
            sign = sign * -1

        # Adding value in sum
        Sum += sign

    # Return Answer
    return Sum

# Driver Code
if __name__ == "__main__":

    N = 6
    print(SeriesSum(N))

    # This code is contributed by ukasp.

C#

// C# program of the above approach
using System;

public class GFG{

// Function to find the sum of
// the series 1! - 2! + 3! - 4!
// + 5!... till the Nth term
static int SeriesSum(int N)
{

    // Stores the required sum
    // and factorial value
    int Sum = 0, fact = 1;

    // Loop to calculate factorial
    // and sum them with their
    // respective sign
    for (int i = 1; i <= N; i++) {

        // Calculate factorial
        fact = fact * i;

        // Stores the sign
        int sign = fact;

        // If position is even sign
        // must be negative
        if (i % 2 == 0) {
            sign = sign * -1;
        }

        // Adding value in sum
        Sum += sign;
    }

    // Return Answer
    return Sum;
}

// Driver Code
public static void Main(String[] args)
{
    int N = 6;
    Console.Write(SeriesSum(N));
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>
// javascript program of the above approach
// Function to find the sum of
// the series 1! - 2! + 3! - 4!
// + 5!... till the Nth term
function SeriesSum(N)
{

    // Stores the required sum
    // and factorial value
    var Sum = 0, fact = 1;

    // Loop to calculate factorial
    // and sum them with their
    // respective sign
    for (var i = 1; i <= N; i++) {

        // Calculate factorial
        fact = fact * i;

        // Stores the sign
        var sign = fact;

        // If position is even sign
        // must be negative
        if (i % 2 == 0) {
            sign = sign * -1;
        }

        // Adding value in sum
        Sum += sign;
    }

    // Return Answer
    return Sum;
}

// Driver Code
var N = 6;
document.write(SeriesSum(N));

// This code is contributed by 29AjayKumar
</script>

Output

-619

Time Complexity: O(N)  
Auxiliary Space: O(1)