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)

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

// 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; 
} 

C

// 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; 
} 

Java

// 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));  
    } 
} 

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#"]

// 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. 

PHP

<?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. 
?> 

Output :

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My Personal Notes

Solving f(n)= (1) + (23) + (45*6) … n using Recursion

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

C

Java

Python3

PHP

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Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

C

Java

Python3

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