Sum of the series 0.6, 0.06, 0.006, 0.0006, ...to n terms

Sum of the series 0.6, 0.06, 0.006, 0.0006, …to n terms

Last Updated : 19 Mar, 2021

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.
?>

Javascript

<script>
// javascript 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 / Math.pow(10, n));
}
 
// Driver Code
let n = 2 ;
   document.write(sumOfSeries(n).toFixed(5)) ;
    
// This code is contributed by aashish1995
 
</script>

Output:

0.65934

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