# Find the sum of N terms of the series 1/1*3, 1/3*5, 1/5*7, ….

• Last Updated : 14 Feb, 2022

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 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 approachimport 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 Codeif __name__ == "__main__":       # Value of N    N = 3       print(findSum(N)) # This code is contributed by Abhishek Thakur.

## C#

 // C# program to implement// the above approachusing 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



Output

0.428571

Time Complexity: O(1)
Auxiliary Space: O(1)

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