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

• Last Updated : 19 Jan, 2022

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

The given G.P. series Here,  Thus, using the sum of G.P. formula for r<1 Substituting the values of a and r in the above equation  Illustration:

Input: n = 3
Output: 0.888
Explanation: 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 using namespace std; // Function to calculate sum of// given series till Nth termdouble sumOfSeries(double N){    return (8 * ((pow(10, N) - 1) / pow(10, N))) / 9;} // Driver codeint main(){    double N = 5;    cout << sumOfSeries(N);    return 0;}

## Java

 // Java program to implement// the above approachimport 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 termdef sumOfSeries(N):    return (8 * (((10 ** N) - 1) / (10 ** N))) / 9; # Driver codeN = 5;print(sumOfSeries(N)); # This code is contributed by gfgking

## C#

 // C# program to implement// the above approachusing 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

 
Output
0.88888

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

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