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Sum of the series 0.6, 0.06, 0.006, 0.0006, …to n terms
  • Last Updated : 03 Aug, 2018

Given the number of terms i.e. n. Find the sum of the series 0.6, 0.06, 0.006, 0.0006, …to n terms.

Examples:

Input : 2
Output : 0.65934

Input : 3
Output : 0.665334

Lets denote the sum by S:
Using the formula S_{n}=\frac{a\left ( 1-r^{n} \right )}{1-r}, we have [since r<1]

S_{n}=\frac{\frac{6}{10}\left \{ 1-\left ( \frac{1}{10}  \right )^n\right \}}{1-\frac{1}{10}}



S_{n}=\frac{6}{9}\left ( 1-\frac{1}{10^n} \right )

S_{n}=\frac{2}{3}\left ( 1-\frac{1}{10^n} \right )

Hence the required sum is S_{n}=\frac{2}{3}\left ( 1-\frac{1}{10^n} \right )

Below is the implementation:

C++




// CPP program to find sum of 0.6, 0.06,
// 0.006, 0.0006, ...to n terms
#include <bits/stdc++.h>
using namespace std;
  
// function which return the
// the sum of series
float sumOfSeries(int n)
{
    return (0.666) * (1 - 1 / pow(10, n));
}
  
// Driver code
int main()
{
    int n = 2;
    cout << sumOfSeries(n);
}


Java




// java program to find sum of 0.6, 0.06,
// 0.006, 0.0006, ...to n terms
import java.io.*;
  
class GFG 
{
    // function which return the
    // the sum of series
    static double sumOfSeries(int n)
    {
        return (0.666) * (1 - 1 /Math. pow(10, n));
    }
      
      
    // Driver code
    public static void main (String[] args) 
    {
        int n = 2;
        System.out.println ( sumOfSeries(n));
          
    }
}
  
// This code is contributed by vt_m


Python3




# Python3 program to find 
# sum of 0.6, 0.06, 0.006, 
# 0.0006, ...to n terms
import math
  
# function which return 
# the sum of series
def sumOfSeries(n):
    return ((0.666) * 
            (1 - 1 / pow(10, n)));
  
# Driver code
n = 2;
print(sumOfSeries(n));
  
# This code is contributed by mits


C#




// C# program to find sum of 0.6, 0.06,
// 0.006, 0.0006, ...to n terms
using System;
  
class GFG {
      
    // function which return the
    // the sum of series
    static double sumOfSeries(int n)
    {
        return (0.666) * (1 - 1 /Math. Pow(10, n));
    }
      
    // Driver code
    public static void Main () 
    {
        int n = 2;
          
        Console.WriteLine( sumOfSeries(n));
          
    }
}
  
// This code is contributed by vt_m


PHP




<?php
// PHP program to find sum of 0.6, 0.06,
// 0.006, 0.0006, ...to n terms
  
// function which return the
// the sum of series
function sumOfSeries($n)
{
    return (0.666) * (1 - 1 / 
                pow(10, $n));
}
  
// Driver code
$n = 2;
echo(sumOfSeries($n));
  
// This code is contributed by Ajit.
?>



Output:

0.65934

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