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Print the first N terms of the series 6, 28, 66, 120, 190, 276, …
  • Difficulty Level : Hard
  • Last Updated : 14 Sep, 2020

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++

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// 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;
}

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Java

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// 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

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Python3

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# Python3 program for the above apporch
  
# 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   

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C#

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// 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

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Output: 

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

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

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