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

  • Last Updated : 15 Mar, 2021

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




// 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;
}

Python




# 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)

Java




// 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);
    }
}

C#




// 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);
    }
}

PHP




<?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);
?>  

Javascript




<script>
 
// javascript implementation of the approach
 
 
// Function to return the nth term of the given series
function Nthterm( n)
{
 
    // nth term
    let numerator = Math.pow(n, 2) - 1;
    let denomenator = 2 * n - 3;
     document.write( numerator + "/" + denomenator);
}
 
// Driver code
let n = 3;
 
    Nthterm(n);
 
// This code contributed by gauravrajput1
 
</script>
Output: 
8/3

 

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