Print first N terms of series (0.25, 0.5, 0.75, …) in fraction representation

Given an integer N, the task is to print the first N terms of the series in their fraction form i.e. 

1/4, 1/2, 3/4, 1, 5/4, …

The above series has values as 0.25, 0.5, 0.75, 1, 1.25, ….etc. It is an Arithmetic progression that begins with 0.25 and has a difference of 0.25.
Examples:

Input: N = 6 
Output: 1/4 1/2 3/4 1 5/4 3/2
Input: N = 9 
Output: 1/4 1/2 3/4 1 5/4 3/2 7/4 2 9/4 

Approach: Consider the first four terms of the series as the base terms. Store the numerator elements and denominator elements separately. 
Consider the first term 1/4, the fifth term is 1 + (1 * 4) / 4 which is 1/5. 
Similarly, consider the second term 1/2 the sixth term is 1 + (1 * 2) / 2 which is 3/2. 
Hence, we can consider the denominators will always be either 2, 4 or no denominator and the numerator of the term can be calculated from the denominator.
Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to print the required series
void printSeries(int n)
{
    // Numerators for the first four numerators
    // of the series
    int nmtr[4] = { 1, 1, 1, 3 };

    // Denominators for the first four denominators
    // of the series
    int dntr[4] = { 0, 4, 2, 4 };

    for (int i = 1; i <= n; i++) {

        // If location of the term in the series is
        // a multiple of 4 then there will be no denominator
        if (i % 4 == 0)
            cout << nmtr[i % 4] + (i / 4) - 1 << " ";

        // Otherwise there will be denominator
        else {

            // Printing the numerator and the denominator terms
            cout << nmtr[i % 4] + ((i / 4) * dntr[i % 4])
                 << "/" << dntr[i % 4] << " ";
        }
    }
}

// Driver code
int main()
{
    int n = 9;
    printSeries(n);
    return 0;
}

Java

// Java implementation of the approach

class GFG
{

// Function to print the required series
public static void printSeries(int n)
{
    // Numerators for the first four numerators
    // of the series
    int[] nmtr = new int[]{ 1, 1, 1, 3 };

    // Denominators for the first four denominators
    // of the series
    int[] dntr = new int[]{ 0, 4, 2, 4 };

    for (int i = 1; i <= n; i++)
    {

        // If location of the term in the series is
        // a multiple of 4 then there will be no denominator
        if (i % 4 == 0)
            System.out.print( nmtr[i % 4] + (i / 4) - 1 + " ");

        // Otherwise there will be denominator
        else
        {

            // Printing the numerator and the denominator terms
            System.out.print( nmtr[i % 4] + ((i / 4) * dntr[i % 4])
                +"/" + dntr[i % 4] +" ");
        }
    }
}

// Driver code
public static void main(String[] args)
{
    int n = 9;
    printSeries(n);
}
}

// This code is contributed
// by 29AjayKumar

Python3

# Python 3 implementation of the approach

# Function to print the required series
def printSeries(n):

    # Numerators for the first four
    # numerators of the series
    nmtr = [1, 1, 1, 3]

    # Denominators for the first four
    # denominators of the series
    dntr = [0, 4, 2, 4]

    for i in range(1, n + 1, 1):

        # If location of the term in the
        # series is a multiple of 4 then
        # there will be no denominator
        if (i % 4 == 0):
            print(nmtr[i % 4] + int(i / 4) - 1,
                                     end = " ")

        # Otherwise there will be denominator
        else:

            # Printing the numerator and
            # the denominator terms
            print(nmtr[i % 4] + (int(i / 4) *
                    dntr[i % 4]), end = "")
            print("/", end = "")
            print(dntr[i % 4], end = " ")

# Driver code
if __name__ == '__main__':
    n = 9
    printSeries(n)

# This code is contributed by
# Shashank_Sharma

C#

// C# implementation of the approach
using System;

class GFG
{

// Function to print the required series
static void printSeries(int n)
{

    // Numerators for the first four numerators
    // of the series
    int[] nmtr = { 1, 1, 1, 3 };

    // Denominators for the first four denominators
    // of the series
    int[] dntr = { 0, 4, 2, 4 };

    for (int i = 1; i <= n; i++)
    {

        // If location of the term in the series is
        // a multiple of 4 then there will be no denominator
        if (i % 4 == 0)
            Console.Write((nmtr[i % 4] + (i / 4) - 1) + " ");

        // Otherwise there will be denominator
        else
        {

            // Printing the numerator and the denominator terms
            Console.Write((nmtr[i % 4] + ((i / 4) * dntr[i % 4])) +
                                        "/" + dntr[i % 4] + " ");
        }
    }
}

// Driver code
public static void Main()
{
    int n = 9;
    printSeries(n);
}
}

// This code is contributed
// by Akanksha Rai

Javascript

<script>
// javascript implementation of the approach

// Function to print the required series
function printSeries( n)
{

    // Numerators for the first four numerators
    // of the series
    let nmtr = [ 1, 1, 1, 3 ];

    // Denominators for the first four denominators
    // of the series
    let dntr = [ 0, 4, 2, 4 ];
    for (let i = 1; i <= n; i++)
    {

        // If location of the term in the series is
        // a multiple of 4 then there will be no denominator
        if (i % 4 == 0)
            document.write( nmtr[i % 4] + (i / 4) - 1 + " ");

        // Otherwise there will be denominator
        else {

            // Printing the numerator and the denominator terms
           document.write( nmtr[i % 4] + (parseInt(i / 4) * dntr[i % 4])
                 + "/" + dntr[i % 4] + " ");
        }
    }
}

// Driver code
    let n = 9;
    printSeries(n);

// This code is contributed by Rajput-Ji

</script>

Output:

1/4 1/2 3/4 1 5/4 3/2 7/4 2 9/4

Time complexity: O(n) for given input n, because using a for loop

Auxiliary space: O(1) It is using constant space