查找系列 0、6、24、60、120 的第 N 项...

给定一个正整数N ，任务是找到序列的第 N 项

0, 6, 24, 60, 120…till N terms

例子：

Input: N = 5
Output: 120

Input: N = 10
Output: 990

方法：

从给定的系列中，找到第 N项的公式-

1st term = 1 ^ 3 – 1 = 0

2nd term = 2 ^ 3 – 2 = 6

3rd term = 3 ^ 3 – 3 = 24

4th term = 4 ^ 3 – 4 = 60

.

.

Nth term = N ^ 3 – N

给定系列的第 N项可以概括为-

T_N = N ^ 3 – N

插图：

Input: N = 10
Output: 990
Explanation:
T_N = N ^ 3 – N
     = 10 ^ 3 – 10
     = 1000 – 10
     = 990

以下是上述方法的实现 -

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function to return
// Nth term of the series
int nth(int n)
{
    return n * n * n - n;
}
 
// Driver code
int main()
{
    int N = 5;
    cout << nth(N) << endl;
    return 0;
}

C

// C program to implement
// the above approach
#include 
 
// Function to return
// Nth term of the series
int nth(int n)
{
    return n * n * n - n;
}
 
// Driver code
int main()
{
    // Value of N
    int N = 5;
    printf("%d", nth(N));
    return 0;
}

Java

// Java program to implement
// the above approach
import java.io.*;
 
class GFG {
    // Driver code
    public static void main(String[] args)
    {
        int N = 5;
        System.out.println(nth(N));
    }
    // Function to return
    // Nth term of the series
    public static int nth(int n)
    {
        return n * n * n - n;
    }
}

Python

# Python program to implement
# the above approach
 
# Function to return
# Nth term of the series
def nth(n):
    return n * n * n - n
 
# Driver code
N = 5
print(nth(N))
 
# This code is contributed by Samim Hossain Mondal.

C#

using System;
 
public class GFG
{
 
  // Function to return
  // Nth term of the series
  public static int nth(int n) { return n * n * n - n; }
 
  // Driver code
  static public void Main()
  {
 
    // Code
    int N = 5;
    Console.Write(nth(N));
  }
}
 
// This code is contributed by Potta Lokesh

Javascript

      // JavaScript code for the above approach
 
      // Function to return
      // Nth term of the series
      function nth(n)
      {
          return n * n * n - n;
      }
 
      // Driver code
      let N = 5;
      document.write(nth(N) + '<br>');
 
     // This code is contributed by Potta Lokesh
  

输出

120

时间复杂度： O(1)
辅助空间： O(1)