Imprime los primeros N términos de la serie 6, 28, 66, 120, 190, 276, …

Dado un número N , la tarea es imprimir los primeros N términos de la serie 6, 28, 66, 120, 190, 276, etc.
Ejemplos:

Entrada: N = 10 
Salida: 6 28 66 120 190 276 378 496 630 780
Entrada: N = 4 
Salida: 6 28 66 120

Enfoque: Para resolver el problema mencionado anteriormente, debemos observar el siguiente patrón:

La fórmula general está dada por: 
k * (2 * k – 1) , donde inicialmente, k = 2

A continuación se muestra la implementación del enfoque anterior:

// 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 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 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# 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

<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>

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

Complejidad temporal: O(N)  
Espacio auxiliar: O(1)