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-

S_{N}=\frac{1}{9}(N-\frac{1}{9}(1-(\frac{1}{10})^{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

\frac{1}{9}(0.9+0.99+0.999+...+N terms)

\frac{1}{9}((1-0.1)+(1-0.11)+(1-0.111)+...+Nterms)

\frac{1}{9}((1+1+1+...+N)-(0.1+0.11+0.111+...+0.1^{N}))

\frac{1}{9}(N-(0.1+0.11+0.111+...+0.1^{N}))                   -(1)

The series 0.1+0.11+0.111+...+0.1^{N}              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-

S_{N}=\frac{a*(1-r^{N})}{1-r}

Substituting the values of a and r in the equation-

S_{N}=\frac{0.1*(1-0.1^{N})}{1-0.1}

S_{N}=\frac{1}{9}(1-(\frac{1}{10})^{N})                      -(2)

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

S_{N}=\frac{1}{9}(N-\frac{1}{9}(1-(\frac{1}{10})^{N}))

Illustration:

Input: N = 6
Output: 0.654321
Explanation:
S_{N}=\frac{1}{9}(N-\frac{1}{9}(1-(\frac{1}{10})^{N}))              
S_{N}=\frac{1}{9}(6-\frac{1}{9}(1-(\frac{1}{10})^{6}))              
S_{N}=\frac{1}{9}(6-0.111111)              
S_{N}=\frac{1}{9}(5.888889)              
S_{N}=0.654321

Below is the implementation of the above approach-

// 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 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.

# 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# 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.

<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>

0.654321

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