Решение f (n) = (1) + (2 * 3) + (4 * 5 * 6)… n с использованием рекурсии

Опубликовано: 7 Января, 2022

Пример :

 Ввод: 2
Выход: 7
Ряд: (1) + (2 * 3)

Ввод: 4
Выход: 5167
Ряды: (1) + (2 * 3) + (4 * 5 * 6) + (7 * 8 * 9 * 10)

Рекомендуется: сначала решите эту проблему на «ПРАКТИКЕ», прежде чем переходить к решению.

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
", 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

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.
?>

Javascript

<script>
 
// JavaScript Program to print the
// solution of the series
// f(n)= (1) + (2*3) + (4*5*6)
// ... n using recursion
 
/**
    * 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
    */
       
    function seriesSum(calculated,
                         current, N)
    {
        let 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
        let N = 5;
       
        // invoking the method
        // to calculate the sum
        document.write(seriesSum(1, 1, N));
     
    // This code is contributed by target_2.
</script>

Выход :

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Решение f (n) = (1) + (2 * 3) + (4 * 5 * 6)… n с использованием рекурсии

C++

C

Java

Python3

C#

PHP

Javascript

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