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Sum of series 8/10, 8/100, 8/1000, 8/10000. . . till N terms

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  • Last Updated : 16 Aug, 2022
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Given a positive integer n, the task is to find the sum of series 

8/10 + 8/100 + 8/1000 + 8/10000. . . till Nth term

Examples:

Input: n = 3
Output: 0.888

Input: n = 5
Output: 0.88888

Approach: 

The total sum till nth term of the given G.P. series can be generalized as-

S_{n}=\frac{8}{9}(1-(\frac{1}{10})^{n})

The above formula can be derived following the series of steps-

The given G.P. series

\frac{8}{10}+\frac{8}{100}+\frac{8}{1000}+\frac{8}{10000}+......              

Here,

a=\frac{8}{10}              

r=\frac{1}{10}              

Thus, using the sum of G.P. formula for r<1

S_{n}=\frac{a(1-r^{n})}{1-r}

Substituting the values of a and r in the above equation

S_{n}=\frac{\frac{8}{10}(1-\frac{1}{10}^{n})}{1-\frac{1}{10}}

S_{n}=\frac{8}{9}(1-(\frac{1}{10})^{n})

Illustration:

Input: n = 3
Output: 0.888
Explanation:
S_{n}=\frac{8}{9}(1-(\frac{1}{10})^{n})
S_{n}=\frac{8}{9}(1-(\frac{1}{10})^{3})
        = 0.888 * 0.999
        = 0.888

Below is the implementation of the above problem-

C++




// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate sum of
// given series till Nth term
double sumOfSeries(double N)
{
    return (8 * ((pow(10, N) - 1) / pow(10, N))) / 9;
}
 
// Driver code
int main()
{
    double N = 5;
    cout << sumOfSeries(N);
    return 0;
}


Java




// Java program to implement
// the above approach
import java.util.*;
public class GFG
{
   
    // Function to calculate sum of
    // given series till Nth term
    static double sumOfSeries(double N)
    {
        return (8
                * ((Math.pow(10, N) - 1) / Math.pow(10, N)))
            / 9;
    }
 
    // Driver code
    public static void main(String args[])
    {
        double N = 5;
        System.out.print(sumOfSeries(N));
    }
}
 
// This code is contributed by Samim Hossain Mondal.


Python3




# Python code for the above approach
 
# Function to calculate sum of
# given series till Nth term
def sumOfSeries(N):
    return (8 * (((10 ** N) - 1) / (10 ** N))) / 9;
 
# Driver code
N = 5;
print(sumOfSeries(N));
 
# This code is contributed by gfgking


C#




// C# program to implement
// the above approach
using System;
class GFG
{
   
    // Function to calculate sum of
    // given series till Nth term
    static double sumOfSeries(double N)
    {
        return (8
                * ((Math.Pow(10, N) - 1) / Math.Pow(10, N)))
            / 9;
    }
 
    // Driver code
    public static void Main()
    {
        double N = 5;
        Console.WriteLine(sumOfSeries(N));
    }
}
 
// This code is contributed by ukasp.


Javascript




<script>
        // JavaScript code for the above approach
 
        // Function to calculate sum of
        // given series till Nth term
        function sumOfSeries(N) {
            return (8 * ((Math.pow(10, N) - 1) / Math.pow(10, N))) / 9;
        }
 
        // Driver code
        let N = 5;
        document.write(sumOfSeries(N));
 
  // This code is contributed by Potta Lokesh
    </script>


Output

0.88888

Time Complexity: O(1)
Auxiliary Space: O(1)


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