Find geometric sum of the series using recursion
Given an integer N, we need to find the geometric sum of the following series using recursion.
1 + 1/3 + 1/9 + 1/27 + … + 1/(3^n)
Examples:
Input N = 5 Output: 1.49794 Input: N = 7 Output: 1.49977
Approach:
In the above-mentioned problem, we are asked to use recursion. We will calculate the last term and call recursion on the remaining n-1 terms each time. The final sum returned is the result.
Below is the implementation of the above approach:
C++
// CPP implementation to Find the // geometric sum of the series using recursion #include <bits/stdc++.h> using namespace std; // function to find the sum of given series double sum( int n) { // base case if (n == 0) return 1; // calculate the sum each time double ans = 1 / ( double ) pow (3, n) + sum(n - 1); // return final answer return ans; } // Driver code int main() { // integer initialisation int n = 5; cout << sum(n) << endl; return 0; } |
Java
// JAVA implementation to Find the // geometric sum of the series using recursion import java.util.*; class GFG { static double sum( int n) { // base case if (n == 0 ) return 1 ; // calculate the sum each time double ans = 1 / ( double )Math.pow( 3 , n) + sum(n - 1 ); // return final answer return ans; } // Driver code public static void main(String[] args) { // integer initialisation int n = 5 ; // print result System.out.println(sum(n)); } } |
Python3
# CPP implementation to Find the # geometric sum of the series using recursion def sum (n): # base case if n = = 0 : return 1 # calculate the sum each time # and return final answer return 1 / pow ( 3 , n) + sum (n - 1 ) n = 5 ; print ( sum (n)); |
C#
// C# implementation to Find the // geometric sum of the series using recursion using System; class GFG { static double sum( int n) { // base case if (n == 0) return 1; // calculate the sum each time double ans = 1 / ( double )Math.Pow(3, n) + sum(n - 1); // return final answer return ans; } // Driver code static public void Main() { int n = 5; Console.WriteLine(sum(n)); } } |
Javascript
<script> // Javascript implementation to Find the // geometric sum of the series using recursion // function to find the sum of given series function sum(n) { // base case if (n == 0) return 1; // calculate the sum each time var ans = 1 / Math.pow(3, n) + sum(n - 1); // return final answer return ans; } // Driver code // integer initialisation var n = 5; document.write( sum(n).toFixed(5)); </script> |
Output:
1.49794
Time Complexity: O(N)
Auxiliary Space: O(N)
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