Find Nth term of the series 0, 6, 24, 60, 120…

Given a positive integer N, the task is to find the Nth term of the series

0, 6, 24, 60, 120…till N terms

Examples:

Input: N = 5
Output: 120

Input: N = 10
Output: 990

Approach: 

From the given series, find the formula for Nth term-

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

The Nth term of the given series can be generalized as-

T_N = N ^ 3 – N

Illustration:

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

Below is the implementation of the above approach-

C++

// C++ program to implement
// the above approach
#include <iostream>
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 <stdio.h>

// 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

<script>
      // 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
  </script>

Output

120

Time Complexity: O(1) // since no loop is used the algorithm takes up constant time to perform the operations
Auxiliary Space: O(1) // since no extra array is used so the space taken by the algorithm is constant