Find Nth term of the series 0, 2, 6, 12, 20, 30, 42...

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

0, 2, 6, 12, 20...till N terms

Examples:

Input: N = 7
Output: 42

Input: N = 10
Output: 90

Approach:

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

1st term = 1 * (1 - 1) = 0

2nd term = 2 * (2 - 1) = 2

3rd term = 3 * (3 - 1) = 6

4th term = 4 * (4 - 1) = 12

.

.

Nth term = N * (N - 1)

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

T_N = N * (N - 1)

Illustration:

Input: N = 7
Output: 42
Explanation:
T_N = N * (N - 1)
     = 7 * (7 - 1)
     = 7 * 6
     = 42

Below is the implementation of the above approach-

// C++ program to implement
// the above approach
#include <iostream>
using namespace std;

// Function to return
// Nth term of the series
int nthTerm(int n)
{
    return n * n - n;
}

// Driver code
int main()
{
    // Value of N
    int N = 7;
    cout << nthTerm(N) << endl;
    return 0;
}

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

// Function to return
// Nth term of the series
int nthTerm(int n)
{
    return n * n - n;
}

// Driver code
int main()
{
    // Value of N
    int N = 7;
    printf("%d", nthTerm(N));
    return 0;
}

// Java program to implement
// the above approach
import java.io.*;

class GFG {
    public static void main(String[] args)
    {
        // Value of N
        int N = 7;
        System.out.println(nthTerm(N));
    }

    // Function to return
    // Nth term of the series
    public static int nthTerm(int n)
    {
        return n * n - n;
    }
}

# Python code for the above approach 

# Function to return
# Nth term of the series
def nthTerm(n):
    return n * n - n;

# Driver code

# Value of N
N = 7;
print(nthTerm(N));

# This code is contributed by Saurabh Jaiswal

using System;

public class GFG 
{

  // Function to return
  // Nth term of the series
  public static int nthTerm(int n)
  {
    return n * n - n;
  }
  
  static public void Main()
  {

    // Code
    // Value of N
    int N = 7;
    Console.Write(nthTerm(N));
  }
}

// This code is contributed by Potta Lokesh

<script>
        // JavaScript code for the above approach 


        // Function to return
        // Nth term of the series
        function nthTerm(n) {
            return n * n - n;
        }

        // Driver code

        // Value of N
        let N = 7;
        document.write(nthTerm(N) + '<br>');

       // This code is contributed by Potta Lokesh
    </script>

42

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