Programa para imprimir las series 1, 9, 17, 33, 49, 73, 97… hasta N términos

Dado un número N , la tarea es imprimir los primeros N términos de la serie:
 

Ejemplos: 
 

Entrada: N = 7 
Salida: 1, 9, 17, 33, 49, 73, 97
Entrada: N = 3 
Salida: 1, 9, 17 
 

Enfoque: De la serie dada, encuentre la fórmula para el N-ésimo término: 
 

1st term = 1
2nd term = 9 = 2 * 4 + 1
3rd term = 17 = 2 * 9 - 1
4th term = 33 = 2 * 16 + 1
5th term = 49 = 2 * 25 - 1
6th term = 73 = 2 * 36 + 1
.
.
Nth term = (2 * N2 + (-1)N)

Por lo tanto: 
 

Enésimo término de la serie 

*** QuickLaTeX no puede compilar la fórmula:
 

*** Mensaje de error:
Error: nada que mostrar, la fórmula está vacía

Luego itere sobre los números en el rango [1, N] para encontrar todos los términos usando la fórmula anterior e imprímalos.
A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of the above approach
 
#include "bits/stdc++.h"
using namespace std;
 
// Function to print the series
void printSeries(int N)
{
 
    int ith_term = 0;
 
    // Generate the ith term and
    // print it
    for (int i = 1; i <= N; i++) {
 
        ith_term = i % 2 == 0
                       ? 2 * i * i + 1
                       : 2 * i * i - 1;
        cout << ith_term << ", ";
    }
}
 
// Driver Code
int main()
{
    int N = 7;
 
    printSeries(N);
    return 0;
}

// Java implementation of the above approach
import java.util.*;
 
class GFG{
  
// Function to print the series
static void printSeries(int N)
{
  
    int ith_term = 0;
  
    // Generate the ith term and
    // print it
    for (int i = 1; i <= N; i++) {
  
        ith_term = i % 2 == 0
                       ? 2 * i * i + 1
                       : 2 * i * i - 1;
        System.out.print(ith_term+ ", ");
    }
}
  
// Driver Code
public static void main(String[] args)
{
    int N = 7;
  
    printSeries(N);
}
}
 
// This code is contributed by PrinciRaj1992

# Python implementation of the above approach
 
# Function to print series
def printSeries(N):
 
    ith_term = 0;
 
    # Generate the ith term and
    # print
    for i in range(1,N+1):
 
        ith_term = 0;
        if(i % 2 == 0):
            ith_term = 2 * i * i + 1;
        else:
            ith_term = 2 * i * i - 1;
        print(ith_term,end= ", ");
     
# Driver Code
if __name__ == '__main__':
    N = 7;
 
    printSeries(N);
     
# This code is contributed by Princi Singh

// C# implementation of the above approach
using System;
 
class GFG{
 
// Function to print the series
static void printSeries(int N)
{
 
    int ith_term = 0;
 
    // Generate the ith term and
    // print it
    for (int i = 1; i <= N; i++) {
 
        ith_term = i % 2 == 0? 2 * i * i + 1:
                                2 * i * i - 1;
        Console.Write(ith_term+ ", ");
    }
}
 
// Driver Code
public static void Main()
{
    int N = 7;
 
    printSeries(N);
}
}
 
// This code is contributed by AbhiThakur

<script>
// javascript implementation of the above approach
 
// Function to print the series
function printSeries( N)
{
    let ith_term = 0;
 
    // Generate the ith term and
    // print it
    for (let i = 1; i <= N; i++)
    {
        ith_term = i % 2 == 0
                       ? 2 * i * i + 1
                       : 2 * i * i - 1;
        document.write( ith_term + ", ");
    }
}
 
// Driver Code
    let N = 7;
 
    printSeries(N);
     
    // This code is contributed by gauravrajput1
 
</script>

1, 9, 17, 33, 49, 73, 97,

Complejidad de tiempo: O(N)

Espacio Auxiliar: O(1)