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++
#include <bits/stdc++.h>
using namespace std;
double sum( int n)
{
if (n == 0)
return 1;
double ans = 1 / ( double ) pow (3, n) + sum(n - 1);
return ans;
}
int main()
{
int n = 5;
cout << sum(n) << endl;
return 0;
}
|
Java
import java.util.*;
class GFG {
static double sum( int n)
{
if (n == 0 )
return 1 ;
double ans = 1 / ( double )Math.pow( 3 , n) + sum(n - 1 );
return ans;
}
public static void main(String[] args)
{
int n = 5 ;
System.out.println(sum(n));
}
}
|
Python3
def sum (n):
if n = = 0 :
return 1
return 1 / pow ( 3 , n) + sum (n - 1 )
n = 5 ;
print ( sum (n));
|
C#
using System;
class GFG {
static double sum( int n)
{
if (n == 0)
return 1;
double ans = 1 / ( double )Math.Pow(3, n) + sum(n - 1);
return ans;
}
static public void Main()
{
int n = 5;
Console.WriteLine(sum(n));
}
}
|
Javascript
<script>
function sum(n)
{
if (n == 0)
return 1;
var ans = 1 / Math.pow(3, n) + sum(n - 1);
return ans;
}
var n = 5;
document.write( sum(n).toFixed(5));
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(N)
Last Updated :
09 Aug, 2021
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