系列 8/10、8/100、8/1000、8/10000 的总和。 . .直到 N 项

给定一个正整数n ，任务是找到序列的和

8/10 + 8/100 + 8/1000 + 8/10000. . . till Nth term

例子：

Input: n = 3
Output: 0.888

Input: n = 5
Output: 0.88888

方法：

给定 GP 系列的第 n项的总和可以概括为-

$S_{n}=\frac{8}{9}(1-(\frac{1}{10})^{n})$

上述公式可以通过以下一系列步骤推导出来——

The given G.P. series

$\frac{8}{10}+\frac{8}{100}+\frac{8}{1000}+\frac{8}{10000}+......$ 

Here,

$a=\frac{8}{10}$ 

$r=\frac{1}{10}$ 

Thus, using the sum of G.P. formula for r<1

$S_{n}=\frac{a(1-r^{n})}{1-r}$

Substituting the values of a and r in the above equation

$S_{n}=\frac{\frac{8}{10}(1-\frac{1}{10}^{n})}{1-\frac{1}{10}}$

$S_{n}=\frac{8}{9}(1-(\frac{1}{10})^{n})$

插图：

Input: n = 3
Output: 0.888
Explanation:
$S_{n}=\frac{8}{9}(1-(\frac{1}{10})^{n})$
S_{n}=\frac{8}{9}(1-(\frac{1}{10})^{3})
        = 0.888 * 0.999
        = 0.888

下面是上述问题的实现——

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function to calculate sum of
// given series till Nth term
double sumOfSeries(double N)
{
    return (8 * ((pow(10, N) - 1) / pow(10, N))) / 9;
}
 
// Driver code
int main()
{
    double N = 5;
    cout << sumOfSeries(N);
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
public class GFG
{
   
    // Function to calculate sum of
    // given series till Nth term
    static double sumOfSeries(double N)
    {
        return (8
                * ((Math.pow(10, N) - 1) / Math.pow(10, N)))
            / 9;
    }
 
    // Driver code
    public static void main(String args[])
    {
        double N = 5;
        System.out.print(sumOfSeries(N));
    }
}
 
// This code is contributed by Samim Hossain Mondal.

Python3

# Python code for the above approach
 
# Function to calculate sum of
# given series till Nth term
def sumOfSeries(N):
    return (8 * (((10 ** N) - 1) / (10 ** N))) / 9;
 
# Driver code
N = 5;
print(sumOfSeries(N));
 
# This code is contributed by gfgking

C#

// C# program to implement
// the above approach
using System;
class GFG
{
   
    // Function to calculate sum of
    // given series till Nth term
    static double sumOfSeries(double N)
    {
        return (8
                * ((Math.Pow(10, N) - 1) / Math.Pow(10, N)))
            / 9;
    }
 
    // Driver code
    public static void Main()
    {
        double N = 5;
        Console.WriteLine(sumOfSeries(N));
    }
}
 
// This code is contributed by ukasp.

Javascript

        // JavaScript code for the above approach
 
        // Function to calculate sum of
        // given series till Nth term
        function sumOfSeries(N) {
            return (8 * ((Math.pow(10, N) - 1) / Math.pow(10, N))) / 9;
        }
 
        // Driver code
        let N = 5;
        document.write(sumOfSeries(N));
 
  // This code is contributed by Potta Lokesh
    

输出

0.88888

时间复杂度： O(1)
辅助空间： O(1)