Given a positive integer n, the task is to find the sum of series
8/10 + 8/100 + 8/1000 + 8/10000. . . till Nth term
Examples:
Input: n = 3
Output: 0.888Input: n = 5
Output: 0.88888
Approach:
The total sum till nth term of the given G.P. series can be generalized as-
The above formula can be derived following the series of steps-
The given G.P. series
Here,
Thus, using the sum of G.P. formula for r<1
Substituting the values of a and r in the above equation
Illustration:
Input: n = 3
Output: 0.888
Explanation:
S_{n}=\frac{8}{9}(1-(\frac{1}{10})^{3})
= 0.888 * 0.999
= 0.888
Below is the implementation of the above problem-
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to calculate sum of// given series till Nth termdouble sumOfSeries(double N){ return (8 * ((pow(10, N) - 1) / pow(10, N))) / 9;}// Driver codeint main(){ double N = 5; cout << sumOfSeries(N); return 0;} |
Java
// Java program to implement// the above approachimport 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 termdef sumOfSeries(N): return (8 * (((10 ** N) - 1) / (10 ** N))) / 9;# Driver codeN = 5;print(sumOfSeries(N));# This code is contributed by gfgking |
C#
// C# program to implement// the above approachusing 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
<script> // 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 </script> |
Output
0.88888
Time Complexity: O(log n)
Auxiliary Space: O(1)
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