Encuentra el término n-ésimo de la serie 3, 9, 21, 41, 71…

Dada una serie matemática como 3, 9, 21, 41, 71… Para un entero n dado, tienes que encontrar el n-ésimo número de esta serie. 
Ejemplos: 
 

Input : n = 4 
Output : 41

Input : n = 2
Output : 9

Nuestra primera tarea para resolver este problema es descifrar la serie. Si observa más de cerca la serie, para un n-ésimo término general, el valor será (Σn ² )+(Σn)+1 , donde

• (Σn ² ) : es la suma de los cuadrados de los primeros n-números naturales .
• (Σn) : es la suma de los primeros n-números naturales .
• 1 es un valor constante simple.

Entonces, para calcular cualquier término n-ésimo de una serie dada, digamos f(n) tenemos: 
f(n) = (Σn ² )+(Σn)+1 
= ( ((n*(n+1)*(2n+ 1))/6) + (n*(n+1)/2) + 1 
= (n ³ + 3n ² + 2n + 3 ) /3
 

// Program to calculate
// nth term of a series
#include <bits/stdc++.h>
using namespace std;
 
// func for calualtion
int seriesFunc(int n)
{
    // for summation of square
    // of first n-natural nos.
    int sumSquare = (n * (n + 1)
                  * (2 * n + 1)) / 6;
 
    // summation of first n natural nos.
    int sumNatural = (n * (n + 1) / 2);
 
    // return result
    return (sumSquare + sumNatural + 1);
}
 
// Driver Code
int main()
{
    int n = 8;   
    cout << seriesFunc(n) << endl;
     
    n = 13;
    cout << seriesFunc(13);
     
    return 0;
}

// Java Program to calculate
// nth term of a series
import java.io.*;
 
class GFG
{
    // func for calualtion
    static int seriesFunc(int n)
    {
        // for summation of square
        // of first n-natural nos.
        int sumSquare = (n * (n + 1)
                        * (2 * n + 1)) / 6;
     
        // summation of first n natural nos.
        int sumNatural = (n * (n + 1) / 2);
     
        // return result
        return (sumSquare + sumNatural + 1);
    }
     
    // Driver Code
    public static void main(String args[])
    {
        int n = 8;
        System.out.println(seriesFunc(n));
         
        n = 13;
        System.out.println(seriesFunc(13));
    }
}
 
// This code is contributed by Nikita Tiwari.

# Program to calculate
# nth term of a series
 
# func for calualtion
def seriesFunc(n):
 
    # for summation of square
    # of first n-natural nos.
    sumSquare = (n * (n + 1) *
                (2 * n + 1)) / 6
 
    # summation of first n
    # natural nos.
    sumNatural = (n * (n + 1) / 2)
 
    # return result
    return (sumSquare + sumNatural + 1)
 
# Driver Code
n = 8
print (int(seriesFunc(n)))
 
n = 13
print (int(seriesFunc(n)))
 
# This is code is contributed by Shreyanshi Arun.

// C# program to calculate
// nth term of a series
using System;
 
class GFG
{
    // Function for calualtion
    static int seriesFunc(int n)
    {
        // For summation of square
        // of first n-natural nos.
        int sumSquare = (n * (n + 1)
                        * (2 * n + 1)) / 6;
     
        // summation of first n natural nos.
        int sumNatural = (n * (n + 1) / 2);
     
        // return result
        return (sumSquare + sumNatural + 1);
    }
     
    // Driver Code
    public static void Main()
    {
        int n = 8;
        Console.WriteLine(seriesFunc(n));
         
        n = 13;
        Console.WriteLine(seriesFunc(13));
    }
}
 
// This code is contributed by vt_m.

<?php
// Program to calculate
// nth term of a series
 
// func for calualtion
function seriesFunc($n)
{
    // for summation of square
    // of first n-natural nos.
    $sumSquare = ($n * ($n + 1)
                * (2 * $n + 1)) / 6;
 
    // summation of first n natural nos.
    $sumNatural = ($n * ($n + 1) / 2);
 
    // return result
    return ($sumSquare + $sumNatural + 1);
}
 
// Driver Code
$n = 8;
echo(seriesFunc($n) . "\n");
     
$n = 13;
echo(seriesFunc($n) . "\n");
 
// This code is contributed by Ajit.
?>

<script>
 
// JavaScript Program to calculate
// nth term of a series
 
    // func for calualtion
    function seriesFunc(n)
    {
        // for summation of square
        // of first n-natural nos.
        let sumSquare = (n * (n + 1)
                        * (2 * n + 1)) / 6;
       
        // summation of first n natural nos.
        let sumNatural = (n * (n + 1) / 2);
       
        // return result
        return (sumSquare + sumNatural + 1);
    }
  
// Driver code
 
        let n = 8;
        document.write(seriesFunc(n) + "<br/>");
           
        n = 13;
        document.write(seriesFunc(13));
 
</script>

Producción : 

241
911

Complejidad temporal : O(1) ya que se realizan operaciones constantes

Espacio Auxiliar: O(1)