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

Given an integer N, the task is to find the Nth 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++

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// 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;
}

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Java

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// 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));
    }
}

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Python3

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# 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))

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C#

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// 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. 

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PHP

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<?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);  
?>

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Output:

512


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