Program to find the Nth term of the series 0, 3/1, 8/3, 15/5……..

Given a number N, the task is to find the Nth term of the following series.

0, 3/1, 8/3, 15/5……

Examples:

Input:  n=4
Output: 15/5

Input: n=3
Output: 8/3

Approach: By clearly examining the series we can find the Tn term for the series and with the help of tn we can find the desired result.

Tn=0 + 3/1 +8/3 +15/5…….
We can see that here odd terms are negative and even terms are positive
Tn=((n2-1)/(2*n-3))

Below is the implementation of the above approach.

CPP

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the nth term of the given series
void Nthterm(int n)
{
  
    // nth term
    int numerator = pow(n, 2) - 1;
    int denomenator = 2 * n - 3;
    cout << numerator << "/" << denomenator;
}
  
// Driver code
int main()
{
    int n = 3;
  
    Nthterm(n);
  
    return 0;
}

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Python

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# Python3 implementation of the approach  
     
# Function to return the nth term of the given series  
def Nthterm(n):  
     
    # nth term  
    numerator = n**2-1
    denomenator = 2 * n-3
      
    print(numerator, "/", denomenator)
     
     
# Driver code  
n = 3
Nthterm(n)

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Java

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// Java implementation of the approach
import java.util.*;
import java.lang.*;
import java.io.*;
  
public class GFG {
  
    // Function to return the nth term of the given series
    static void NthTerm(int n)
    {
        int numerator
            = ((int)Math.pow(n, 2)) - 1;
        int denomeanator = 2 * n - 3;
        System.out.println(numerator + "/" + denomeanator);
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int n = 3;
        NthTerm(n);
    }
}

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C#

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// C# implementation of the approach
using System;
public class GFG {
  
    // Function to return the nth term of the given series
    static void NthTerm(int n)
    {
  
        int numerator
            = ((int)Math.Pow(n, 2)) - 1;
        int denomeanator = 2 * n - 3;
        Console.WriteLine(numerator + "/" + denomeanator);
    }
  
    // Driver code
    public static void Main()
    {
        int n = 3;
        NthTerm(n);
    }
}

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PHP

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<?php 
// PHP implementation of the approach 
  
// Function to return the nth term of the given series 
function Nthterm($n
  
    $numerator = (pow($n, 2)) -1;
    $denomenator=2*$n-3;
  
    echo $numerator, "/", $denomenator;
    return $Tn
                      
      
// Driver code 
$n = 3; 
Nthterm($n); 
?>   

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Output:

8/3


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