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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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);   
?> 

Output:

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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