求数列 0，8，64，216，512，.。。

原文:https://www . geesforgeks . org/find-the-n-term-of-series-0-8-64-216-512/

给定一个整数 N ，任务是找到以下系列的 N ^第T5】项:

0, 8, 64, 216, 512, 1000, 1728, .。。

例:

输入: N = 6 输出: 1000 输入: N = 5 输出: 512

进场:

• 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 0、2、4、6、10、… 在 AP 中，使用公式项= a₁+(n–1) d可以找到本系列的第 n 项，其中a₁T13】为第一项， n 为**项位置【T19*
• 要得到原系列中的术语，术语=术语*(术语 ² ) 即术语 ³ 。
• 最后打印术语。

以下是上述方法的实现:

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 语言(一种计算机语言，尤用于创建网站)

// 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));
    }
}

Python 3

# 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.

服务器端编程语言（Professional Hypertext Preprocessor 的缩写）

<?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); 
?>

java 描述语言

<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

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