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++

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// 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;
}

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Java

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// 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));
    }
}

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Python3

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# 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));

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C#

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// 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));
    }
}

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Output:

1.49794

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