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)