Print the first N terms of the series 6, 28, 66, 120, 190, 276, …

Given a number N, the task is to print the first N terms of the series 6, 28, 66, 120, 190, 276, and so on.
Examples:

Input: N = 10 
Output: 6 28 66 120 190 276 378 496 630 780
Input: N = 4 
Output: 6 28 66 120

Approach: To solve the problem mentioned above, we have to observe the below pattern:

The general formula is given by: 
k * (2 * k – 1), where initially, k = 2

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to print the series
void printSeries(int n)
{
    // Initialise the value of k with 2
    int k = 2;

    // Iterate from 1 to n
    for (int i = 0; i < n; i++) {

        // Print each number
        cout << (k * (2 * k - 1))
             << " ";

        // Increment the value of
        // K by 2 for next number
        k += 2;
    }

    cout << endl;
}

// Driver Code
int main()
{
    // Given number N
    int N = 12;

    // Function Call
    printSeries(N);
    return 0;
}

Java

// Java program for the above approach
class GFG{

// Function to print the series
static void printSeries(int n)
{
    // Initialise the value of k with 2
    int k = 2;

    // Iterate from 1 to n
    for (int i = 0; i < n; i++)
    {

        // Print each number
        System.out.print(k * (2 * k - 1) + " ");

        // Increment the value of
        // K by 2 for next number
        k += 2;
    }

    System.out.println();
}

// Driver code
public static void main(String args[])
{
    // Given number N
    int N = 12;

    // Function Call
    printSeries(N);
}
}

// This code is contributed by shivaniisnghss2110

Python3

# Python3 program for the above approach

# Function to print the series
def PrintSeries(n):

    # Initialise the value of k with 2
    k = 2

    # Iterate from 1 to n
    for i in range(0, n):

        # Print each number
        print(k * (2 * k - 1), end = ' ')

        # Increment the value of
        # K by 2 for next number
        k = k + 2

# Driver code

# Given number
n = 12

# Function Call
PrintSeries(n)

# This code is contributed by poulami21ghosh

C#

// C# program for the above approach
using System;

class GFG{

// Function to print the series
static void printSeries(int n)
{

    // Initialise the value of k with 2
    int k = 2;

    // Iterate from 1 to n
    for(int i = 0; i < n; i++)
    {

        // Print each number
        Console.Write(k * (2 * k - 1) + " ");

        // Increment the value of
        // K by 2 for next number
        k += 2;
    }
    Console.WriteLine();
}

// Driver code
public static void Main()
{

    // Given number N
    int N = 12;

    // Function call
    printSeries(N);
}
}

// This code is contributed by sanjoy_62

Javascript

<script>
// javascript program for the above approach

// Function to print the series
function printSeries( n)
{

    // Initialise the value of k with 2
    let k = 2;

    // Iterate from 1 to n
    for (let i = 0; i < n; i++) {

        // Print each number
        document.write((k * (2 * k - 1))
             + " ");

        // Increment the value of
        // K by 2 for next number
        k += 2;
    }

   document.writeln("<br/>");
}

// Driver Code

    // Given number N
    let N = 12;

    // Function Call
    printSeries(N);

// This code is contributed by Rajput-Ji

</script>

Output:

6 28 66 120 190 276 378 496 630 780 946 1128

Time Complexity: O(N)
Auxiliary Space: O(1)