Given a positive integer, N. Find the sum of the first N term of the series-
1/1*3, 1/3*5, 1/5*7, ….
Examples:
Input: N = 3
Output: 0.428571
Input: N = 1
Output: 0.333333
Approach: The sequence is formed by using the following pattern. For any value N-
SN = N / (2 * N + 1)
Below is the implementation of the above approach:
C++
// 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) {
return ( double )N / (2 * N + 1);
} // Driver Code int main()
{ int N = 3;
cout << findSum(N);
} |
Java
// JAVA program to implement // the above approach import java.util.*;
class GFG
{ // Function to return sum of
// N term of the series
public static double findSum( int N)
{
return ( double )N / ( 2 * N + 1 );
}
// Driver Code
public static void main(String[] args)
{
int N = 3 ;
System.out.print(findSum(N));
}
} // This code is contributed by Taranpreet |
Python3
# Python 3 program for the above approach # Function to return sum of # N term of the series def findSum(N):
return N / ( 2 * N + 1 )
# Driver Code if __name__ = = "__main__" :
# Value of N
N = 3 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
public static double findSum( int N)
{
return ( double )N / (2 * N + 1);
}
// Driver Code
public static void Main()
{
int N = 3;
Console.Write(findSum(N));
}
} // This code is contributed by gfgking |
Javascript
<script> // Javascript program to implement // the above approach // Function to return sum of // N term of the series function findSum(N) {
return N / (2 * N + 1);
} // Driver Code let N = 3; document.write(findSum(N)); // This code is contributed by Palak Gupta </script> |
Output
0.428571
Time Complexity: O(1), since there is no loop or recursion.
Auxiliary Space: O(1), since no extra space has been taken.