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

• Difficulty Level : Basic
• Last Updated : 30 Mar, 2023

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 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

 

## 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 = 3 print("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 = 3 sum = 2*((1 - 1 / math.pow(10, n)))/3 print("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)

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