Solving f(n)= (1) + (2*3) + (4*5*6) … n using Recursion

 
Example : 

Input : 2
Output: 7
Series: (1) + (2*3)

Input : 4
Output: 5167
Series: (1) + (2*3) + (4*5*6) + (7*8*9*10)


C++

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// CPP Program to print the solution
// of the series f(n)= (1) + (2*3)
// + (4*5*6) ... n using recursion
#include<bits/stdc++.h>
using namespace std;
  
// Recursive function for 
// finding sum of series
// calculated - number of terms till 
//             which sum of terms has 
//             been calculated
// current - number of terms for which 
//             sum has to becalculated
// N         - Number of terms in the 
//             function to be calculated
int seriesSum(int calculated, int current, 
                            int N) 
{
    int i, cur = 1;
  
    // checking termination condition
    if (current == N + 1) 
        return 0;
  
    // product of terms till current
    for (i = calculated; i < calculated + 
                            current; i++)
        cur *= i; 
  
    // recursive call for adding
    // terms next in the series
    return cur + seriesSum(i, current + 1, N); 
}
  
// Driver Code
int main()
{
    // input number of terms in the series
    int N = 5; 
  
    // invoking the function to
    // calculate the sum
    cout<<seriesSum(1, 1, N)<<endl; 
  
    return 0;
}

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C














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// C Program to print the solution 


// of the series f(n)= (1) + (2*3) 


// + (4*5*6) ... n using recursion 


#include <stdio.h> 


  


// Recursive function for  


// finding sum of series 


// calculated - number of terms till  


//                which sum of terms has  


//              been calculated 


// current - number of terms for which  


//             sum has to becalculated 


// N         - Number of terms in the  


//             function to be calculated 


int seriesSum(int calculated, int current,  


                              int N)  


{ 


    int i, cur = 1; 


  


    // checking termination condition 


    if (current == N + 1)  


        return 0; 


  


    // product of terms till current 


    for (i = calculated; i < calculated +  


                            current; i++) 


        cur *= i;  


  


    // recursive call for adding 


    // terms next in the series 


    return cur + seriesSum(i, current + 1, N);  


} 


  


// Driver Code 


int main() 


{ 


    // input number of terms in the series 


    int N = 5;  


  


    // invoking the function to 


    // calculate the sum 


    printf("%d\n", seriesSum(1, 1, N));  


  


    return 0; 


} 



















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Java














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// Java Program to print the 


// solution of the series 


// f(n)= (1) + (2*3) + (4*5*6)  


// ... n using recursion  


  


class GFG 


{ 


      


    /** 


    * Recursive method for finding  


    * sum of series 


    * 


    * @param calculated number of terms  


    * till which sum of terms has been  


    * calculated @param current number of 


    * terms for which sum has to be calculated. 


    * @param N Number of terms in the function  


    * to be calculated @return sum 


    */


      


    static int seriesSum(int calculated,  


                         int current,  


                         int N)  


    { 


        int i, cur = 1; 


      


        // checking termination condition 


        if (current == N + 1)  


            return 0; 


      


        // product of terms till current 


        for (i = calculated; i < calculated +  


                                current; i++) 


            cur *= i;  


      


        // recursive call for adding  


        // terms next in the series 


        return cur + seriesSum(i, current + 1, N);  


    } 


      


    // Driver Code 


    public static void main(String[] args)  


    { 


        // input number of  


        // terms in the series 


        int N = 5;  


      


        // invoking the method  


        // to calculate the sum 


        System.out.println(seriesSum(1, 1, N));  


    } 


} 



















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Python3





# Python3 Program to print the solution

# of the series f(n)= (1) + (2*3) + (4*5*6)

# … n using recursion


# Recursive function for finding sum of series

# calculated – number of terms till

#		       which sum of terms

#		       has been calculated

# current – number of terms for

#		    which sum has to be

#		    calculated

# N	 – Number of terms in the

#	   function to be calculated

def seriesSum(calculated, current, N): 


	i = calculated;

	cur = 1;


	# checking termination condition

	if (current == N + 1):

		return 0;


	# product of terms till current

	while (i < calculated + current):
		cur *= i;
		i += 1;

	# recursive call for adding
	# terms next in the series
	return cur + seriesSum(i, current + 1, N);

# Driver code

# input number of terms in the series
N = 5; 

# invoking the function
# to calculate the sum
print(seriesSum(1, 1, N)); 

# This code is contributed by mits



[tabby title="C#"]

                











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// C# Program to print the  


// solution of the series  


// f(n)= (1) + (2*3) + (4*5*6) 


// ... n using recursion  


using System; 


  


class GFG  


{ 


      


    // Recursive function for  


    // finding sum of series 


    // calculated - number of terms till   


    //                which sum of terms 


    //              has been calculated 


    // current    - number of terms for which  


    //                sum has to be calculated 


    // N         - Number of terms in the   


    //               function to be calculated 


    static int seriesSum(int calculated,  


                         int current,  


                         int N)  


    { 


          


        int i, cur = 1; 


      


        // checking termination condition 


        if (current == N + 1)  


            return 0; 


      


        // product of terms till current 


        for (i = calculated; i < calculated +  


                                current; i++) 


            cur *= i;  


      


        // recursive call for adding terms 


        // next in the series 


        return cur + seriesSum(i, current + 1, N);  


    } 


      


    // Driver Code 


    public static void Main()  


    { 


          


        // input number of terms 


        // in the series 


        int N = 5;  


      


        // invoking the method to 


        // calculate the sum 


        Console.WriteLine(seriesSum(1, 1, N));  


    } 


} 


  


// This code is contributed by vt_m. 



















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PHP














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<?php 


// PHP Program to print the  


// solution of the series 


// f(n)= (1) + (2*3) + (4*5*6)  


// ... n using recursion 


  


// Recursive function for  


// finding sum of series 


// calculated - number of terms till  


//                which sum of terms  


//              has been calculated 


// current -    number of terms for  


//                which sum has to be  


//              calculated 


// N         - Number of terms in the  


//             function to be calculated 


function seriesSum($calculated, $current, $N)  


{ 


    $i; $cur = 1; 


  


    // checking termination condition 


    if ($current == $N + 1)  


        return 0; 


  


    // product of terms till current 


    for ($i = $calculated; $i < $calculated +  


                              $current; $i++) 


        $cur *= $i;  


  


    // recursive call for adding 


    // terms next in the series 


    return $cur + seriesSum($i, $current + 1, $N);  


} 


  


// Driver code 


  


// input number of  


// terms in the series 


$N = 5;  


  


// invoking the function 


// to calculate the sum 


echo(seriesSum(1, 1, $N));  


  


// This code is contributed by Ajit. 


?> 



















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Output :


365527 


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