Find the sum of N terms of the series 0.1, 0.11, 0.111, …

Given a positive integer, N. Find the sum of the first N term of the series-

 0.1, 0.11, 0.111, 0.1111, ….till N terms

Examples:

Input: N = 6
Output: 0.654321

Input: N = 1
Output: 0.1

Approach:

1st term = 0.1

2nd term = 0.11

3rd term = 0.111

4th term = 0.1111

.

.

Nth term = 1/9(1 – (1/10) ^ N)

The sequence is formed by using the following pattern. For any value N-

Derivation:

The following series of steps can be used to derive the formula to find the sum of N terms-

The series 0.1, 0.11, 0.111, …till N terms can be written as

     -(1)

The series  is in G.P. with

First term a = 0.1  = 10^-1

Common Ratio r = 10^-1

Sum of G.P. for r<1 can be expressed as-

Substituting the values of a and r in the equation-

        -(2)

Substituting the equation (2) in (1), we get-

Illustration:

Input: N = 6
Output: 0.654321
Explanation:

Below is the implementation of the above approach-

C++

// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to return sum of
// N term of the series
double findSum(int N)
{
    int a = pow(10, N);
    return (double)(N * 9 * a - a + 1)
           / (81 * a);
}

// Driver Code
int main()
{
    int N = 6;
    cout << findSum(N);
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;

class GFG {

// Function to return sum of
// N term of the series
static double findSum(double N)
{
    double a = Math.pow(10, N);
    return (double)(N * 9 * a - a + 1)
           / (81 * a);
}

// Driver Code
    public static void main (String[] args) {
          double N = 6;
        System.out.print(findSum(N));
    }
}

// This code is contributed by hrithikgarg03188.

Python3

# Python 3 program for the above approach

# Function to return sum of
# N term of the series
def findSum(N):
    a = pow(10, N)
    return (N * 9 * a - a + 1) / (81 * a)

# Driver Code
if __name__ == "__main__":

    # Value of N
    N = 6
    print(findSum(N))

# This code is contributed by Abhishek Thakur.

C#

// C# program to implement
// the above approach
using System;
class GFG
{

  // Function to return sum of
  // N term of the series
  static double findSum(int N)
  {
    int a = (int)Math.Pow(10, N);
    return (double)(N * 9 * a - a + 1)
      / (81 * a);
  }

  // Driver Code
  public static void Main()
  {
    int N = 6;
    Console.Write(findSum(N));
  }
}

// This code is contributed by Samim Hossain Mondal.

Javascript

<script>
        // JavaScript code for the above approach

        // Function to return sum of
        // N term of the series
        function findSum(N) {
            let a = Math.pow(10, N);
            return (N * 9 * a - a + 1)
                / (81 * a);
        }

        // Driver Code
        let N = 6;
        document.write(findSum(N));

       // This code is contributed by Potta Lokesh
    </script>

Output

0.654321

Time Complexity: O(logN) because it is using inbuilt pow function
Auxiliary Space: O(1)