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

// 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 denominator = 2 * n - 3;
    cout << numerator << "/" << denominator;
}

// 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
    denominator = 2 * n-3

    print(numerator, "/", denominator)


# 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 denominator = 2 * n - 3;
        System.out.println(numerator + "/" + denominator);
    }

    // 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 denominator = 2 * n - 3;
        Console.WriteLine(numerator + "/" + denominator);
    }

    // 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;
    $denominator=2*$n-3;

    echo $numerator, "/", $denominator;
    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 denominator = 2 * n - 3;
     document.write( numerator + "/" + denominator);
}

// Driver code
let n = 3;

    Nthterm(n);

// This code contributed by gauravrajput1

</script>

Output:

8/3

Time Complexity: O(log n) since time complexity of inbuilt pow function is logn

Auxiliary Space: O(1)