Find the nth term of the series 0, 8, 64, 216, 512, . . .

Given an integer N, the task is to find the N^th term of the following series: 

0, 8, 64, 216, 512, 1000, 1728, . . . 

Examples:

Input: N = 6 
Output: 1000
Input: N = 5 
Output: 512 

Approach:

• Given series 0, 8, 64, 216, 512, 1000, 1728, … can also be written as 0 * (0²), 2 * (2²), 4 * (4²), 6 * (6²), 8 * (8²), 10 * (10²), …
• Observe that 0, 2, 4, 6, 10, … is in AP and the nth term of this series can be found using the formula term = a₁ + (n – 1) * d where a₁ is the first term, n is the term position and d is the common difference.
• To get the term in the original series, term = term * (term²) i.e. term³.
• Finally print the term.

Below is the implementation of the above approach:

C++

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

// Function to return the nth term of the given series
long term(int n)
{
    // Common difference
    int d = 2;

    // First term
    int a1 = 0;

    // nth term
    int An = a1 + (n - 1) * d;

    // nth term of the given series
    An = pow(An, 3);
    return An;
}

// Driver code
int main()
{
    int n = 5;

    cout << term(n);

    return 0;
}

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 int nthTerm(int n)
    {

        // Common difference and first term
        int d = 2, a1 = 0;

        // nth term
        int An = a1 + (n - 1) * d;

        // nth term of the given series
        return (int)Math.pow(An, 3);
    }

    // Driver code
    public static void main(String[] args)
    {
        int n = 5;
        System.out.println(nthTerm(n));
    }
}

Python3

# Python3 implementation of the approach

# Function to return the nth term of the given series
def term(n):

    # Common difference
    d = 2

    # First term
    a1 = 0

    # nth term
    An = a1 +(n-1)*d

    # nth term of the given series
    An = An**3
    return An;


# Driver code
n = 5
print(term(n))

C#

// C# implementation of the approach
using System;
public class GFG {

    // Function to return the nth term of the given series
    static int nthTerm(int n)
    {

        // Common difference and first term
        int d = 2, a1 = 0;

        // nth term
        int An = a1 + (n - 1) * d;

        // nth term of the given series
        return (int)Math.Pow(An, 3);
    }

    // Driver code
    public static void Main()
    {
        int n = 5;
        Console. WriteLine(nthTerm(n));
    }
}
// This code is contributed by Mutual singh.

PHP

<?php
// PHP implementation of the approach

// Function to return the nth term of the given series
function term($n)
{

    // Common difference
    $d = 2;

    // First term
    $a1 = 0;

    // nth term
    $An=$a1+($n-1)*$d;

    // nth term of the given series
    return pow($An, 3);

}

// Driver code
$n = 5;
echo term($n);
?>

Javascript

<script>

// javascript implementation of the approach

// Function to return the nth term of the given series
function term(n)
{

    // Common difference
    let d = 2;

    // First term
    let a1 = 0;

    // nth term
    An=a1+(n-1)*d;

    // nth term of the given series
    return Math.pow(An, 3);

}

// Driver code
let n = 5;
document.write( term(n));

// This code is contributed by sravan kumar

</script>

Output:

512

Time Complexity: O(1)

Auxiliary Space: O(1)