找出系列0、8、64、216、512，…的n个项。 。 。

给定整数N ，任务是找到以下系列的^第N^个项：

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

例子：

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

方法：

• 给定系列0、8、64、216、512、1000、1728，…也可以写成0 *(0 ² )，2 *(2 ² )，4 *(4 ² )，6 *(6 ² )， 8 *(8 ² )，10 *(10 ² )，…
• 观察到0，2，4，6，10，…是在AP和该序列的第n项可使用下式术语中找到= ₁ +(N – 1)* d，其中a ₁是第一项，正是位置项， d是共同的区别。
• 要获得原始系列中的术语， term =术语*(term ² )，即term ³ 。
• 最后打印该术语。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
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.

Javascript

 
// 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
 

输出：

512