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

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

Let’s denote the sum by S:
Using the formula , we have [since r<1]

Hence the required sum is
Below is the implementation:

## C++

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

## Java

 // java program to find sum of 0.6, 0.06,// 0.006, 0.0006, ...to n termsimport 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 termsimport math # function which return# the sum of seriesdef sumOfSeries(n):    return ((0.666) *            (1 - 1 / pow(10, n))); # Driver coden = 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 termsusing 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

 

## Javascript

 

Output

0.65934

Method2 :

## C++

 #include #include  double sum_of_series(int n) {    return 0.6 * (1 - pow(1/10.0, n)) / (1 - 1/10.0);} int main() {    int n;    n=3;    std::cout << "Sum of the series to " << n << " terms: " << sum_of_series(n) << std::endl;    return 0;}

## Java

 public class Main {          public static void main(String[] args)          {          int n = 3;          System.out.println("Sum of the series to " + n          + " terms: " + sumOfSeries(n));          }          public static double sumOfSeries(int n)          {          return 0.6 * (1 - Math.pow(1 / 10.0, n))          / (1 - 1 / 10.0);          }     }}

## Python3

 import math def sum_of_series(n):    return 0.6 * (1 - pow(1/10.0, n)) / (1 - 1/10.0) n = 3print("Sum of the series to", n, "terms:", sum_of_series(n))

## C#

 using System; public class Program {    public static double SumOfSeries(int n) {        return 0.6 * (1 - Math.Pow(1.0 / 10, n)) / (1 - 1.0 / 10);    }     public static void Main() {        int n = 3;        Console.WriteLine("Sum of the series to {0} terms: {1}", n, SumOfSeries(n));    }}

## Javascript

 function sumOfSeries(n) {  return 0.6 * (1 - Math.pow(1 / 10, n)) / (1 - 1 / 10);} let n = 3;console.log(Sum of the series to ${n} terms:${sumOfSeries(n)});

Output

Sum of the series to 3 terms: 0.666

Time complexity: O(logn) because using inbuilt pow function
Auxiliary Space: O(1)

### Method3:

1. This code uses the formula for the sum of the series, which is S = 2*((1 – 1/10^n))/3.

2. The code initializes an integer variable n to 3, which represents the number of terms in the series to sum. It then computes the sum of the series using the formula and assigns it to a float variable sum.

3. Finally, the code prints the sum to the console.

Note that we use the math.pow() function from the math module to compute the value of 10^n, since the ** operator in Python does not work with large exponents.

## Python3

 import math n = 3sum = 2*((1 - 1 / math.pow(10, n)))/3print("Sum of the series to", n, "terms:", sum)

## Java

 import java.lang.Math; public class Main {  public static void main(String[] args) {    int n = 3;    double sum = 2*((1 - 1 / Math.pow(10, n)))/3;    System.out.println("Sum of the series to " + n + " terms: " + sum);  }}

## C++

 #include #include  using namespace std; int main() {    int n = 3;    double sum = 2 * ((1 - 1 / pow(10, n))) / 3;    cout << "Sum of the series to " << n << " terms: " << sum << endl;    return 0;}

## C#

 using System; class Program {    static void Main(string[] args) {        int n = 3;        double sum = 2 * ((1 - 1 / Math.Pow(10, n))) / 3;        Console.WriteLine("Sum of the series to {0} terms: {1}", n, sum);    }}

## Javascript

 let n = 3;let sum = 2 * ((1 - 1 / Math.pow(10, n))) / 3;console.log(Sum of the series to ${n} terms:${sum});

Output

Sum of the series to 3 terms: 0.666

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