Given the value of n, find the sum of the series (2 / 3) – (4 / 5) + (6 / 7) – (8 / 9) + – – – – – – – upto n terms.
Examples :
Input : n = 5
Output : 0.744012
Series : (2 / 3) - (4 / 5) + (6 / 7) - (8 / 9) + (10 / 11)
Input : n = 7
Output : 0.754268
Series : (2 / 3) - (4 / 5) + (6 / 7) - (8 / 9) +
(10 / 11) - (12 / 13) + (14 / 15)
C++
// C++ program to find // sum of given series #include <bits/stdc++.h> using namespace std; // Function to find sum of series // up-to n terms double seriesSum(int n) { // initializing counter by 1 int i = 1; // variable to calculate result double res = 0.0; bool sign = true; // while loop until nth term // is not reached while (n > 0) { n--; // boolean type variable // for checking validation if (sign) { sign = !sign; res = res + (double)++i / ++i; } else { sign = !sign; res = res - (double)++i / ++i; } } return res; } // Driver Code int main() { int n = 5; cout << seriesSum(n); return 0; } |
Java
// Java program to find // sum of given series import java.io.*; class GFG { // Function to find sum of series // up-to n terms static double seriesSum(int n) { // initializing counter by 1 int i = 1; // variable to calculate result double res = 0.0; boolean sign = true; // while loop until nth term // is not reached while (n > 0) { n--; // boolean type variable // for checking validation if (sign) { sign = !sign; res = res + (double)++i / ++i; } else { sign = !sign; res = res - (double)++i / ++i; } } return res; } // Driver Code public static void main (String[] args) { int n = 5; System.out.print(seriesSum(n)); } } // This code is contributed by vt_m |
Python3
# Python3 program to find # sum of given series # Function to find # sum of series # up-to n terms def seriesSum(n): # initializing # counter by 1 i = 1; # variable to # calculate result res = 0.0; sign = True; # while loop until nth # term is not reached while (n > 0): n = n - 1; # boolean type variable # for checking validation if (sign): sign = False; res = res + (i + 1) / (i + 2); i = i + 2; else: sign = True; res = res - (i + 1) / (i + 2); i = i + 2; return res; # Driver Code n = 5; print(round(seriesSum(n), 6)); # This code is contributed # by mits |
C#
// C# program to find // sum of given series using System; class GFG { // Function to find sum of // series up-to n terms static double seriesSum(int n) { // initializing counter by 1 int i = 1; // variable to calculate result double res = 0.0; bool sign = true; // while loop until nth term // is not reached while (n > 0) { n--; // boolean type variable // for checking validation if (sign) { sign = !sign; res = res + (double)++i / ++i; } else { sign = !sign; res = res - (double)++i / ++i; } } return res; } // Driver Code public static void Main () { int n = 5; Console.Write(seriesSum(n)); } } // This code is contributed by vt_m |
PHP
<?php // PHP program to find // sum of given series // Function to find sum of series // up-to n terms function seriesSum($n) { // initializing counter by 1 $i = 1; // variable to calculate result $res = 0.0; $sign = true; // while loop until nth term // is not reached while ($n > 0) { $n--; // boolean type variable // for checking validation if ($sign) { $sign = !$sign; $res = $res + (double)++$i / ++$i; } else { $sign = !$sign; $res = $res - (double)++$i / ++$i; } } return $res; } // Driver Code $n = 5; echo(seriesSum($n)); // This code is contributed by Ajit. ?> |
Javascript
<script> // javascript program to find // sum of given series // Function to find sum of series // up-to n terms function seriesSum( n) { // initializing counter by 1 let i = 1; // variable to calculate result let res = 0.0; let sign = true; // while loop until nth term // is not reached while (n > 0) { n--; // boolean type variable // for checking validation if (sign) { sign = !sign; res = res + ++i / ++i; } else { sign = !sign; res = res - ++i / ++i; } } return res; } // Driver Code let n = 5 ; document.write(seriesSum(n).toFixed(6)) ; // This code contributed by aashish1995 </script> |
Output :
0.744012
Time Complexity: O(n), where n represents the given integer.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
