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)