求系列 0.1, 0.11, 0.111, ... 的 N 项之和

给定一个正整数N 。找到系列的前N 项的总和-

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

例子：

Input: N = 6
Output: 0.654321

Input: N = 1
Output: 0.1

方法：

1st term = 0.1

2nd term = 0.11

3rd term = 0.111

4th term = 0.1111

.

.

Nth term = 1/9(1 – (1/10) ^ N)

该序列是通过使用以下模式形成的。对于任何值 N-

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

推导：

以下一系列步骤可用于推导公式以找到 N 项的总和 -

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 eqution (2) in (1), we get-

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

插图：

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

以下是上述方法的实现 -

C++

// C++ program to implement
// the above approach
#include 
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

        // 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
    

输出

0.654321

时间复杂度： O(1)
辅助空间： O(1)