求系列 1/1*3、1/3*5、1/5*7、……的 N 项之和。

给定一个正整数N 。找到系列的前N 项的总和-

1/1*3, 1/3*5, 1/5*7, ….

例子：

Input: N = 3

Output: 0.428571

Input: N = 1

Output: 0.333333

方法：使用以下模式形成序列。对于任何值 N-

S_N = N / (2 * N + 1)

下面是上述方法的实现：

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 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

// 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

输出

0.428571

时间复杂度：O(1)
辅助空间：O(1)