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++
// 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 |
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 :
365527
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