Encuentra el número N de la serie 1, 6, 15, 28, 45, …..

Dados los dos primeros números de una serie. La tarea es encontrar el número N-ésimo (N puede ser hasta 10^18) de esa serie. 
Nota: Cada elemento en la array es dos menos que la media del número anterior y posterior. La respuesta puede ser muy grande, así que imprima la respuesta en el módulo 10^9+9.
Ejemplos : 
 

Input: N = 3
Output: 15
(1 + 15)/2 - 2 = 6

Input: N = 4
Output: 28
(6 + 28)/2 - 2 = 15

Observación: Según el enunciado, la serie formada será 1, 6, 15, 28, 45….. Entonces, la fórmula para el N-ésimo término será: 
 

2*n*n - n

// CPP program to find Nth term of the series
#include <bits/stdc++.h>
using namespace std;
#define mod 1000000009
 
// function to return nth term of the series
int NthTerm(long long n)
{
    long long x = (2 * n * n) % mod;
    return (x - n + mod) % mod;
}
 
// Driver code
int main()
{
    long long N = 4;
 
    // function call
    cout << NthTerm(N);
 
    return 0;
}

// C program to find Nth term of the series
#include <stdio.h>
 
#define mod 1000000009
 
// function to return nth term of the series
int NthTerm(long long n)
{
    long long x = (2 * n * n) % mod;
    return (x - n + mod) % mod;
}
 
// Driver code
int main()
{
    long long N = 4;
 
    // function call
    printf("%d",NthTerm(N));
 
    return 0;
}
 
// This code is contributed by kothavvsaakash.

// Java program to find N-th
// term of the series:
 
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG {
 
    // function to return nth term of the series
    static long NthTerm(long n)
    {
        long x = (2 * n * n) % 1000000009;
        return (x - n + 1000000009) % 1000000009;
    }
 
    // Driver Code
    public static void main(String args[])
    {
 
        // Taking  n as 6
        long N = 4;
 
        // Printing the nth term
        System.out.println(NthTerm(N));
    }
}

# Python 3 program to find 
# N-th term of the series: 
   
 
# Function for calculating 
# Nth term of series 
def NthTerm(N) : 
   
    # return nth term
    x = (2 * N*N)% 1000000009
    return ((x - N + 1000000009)% 1000000009) 
   
# Driver code 
if __name__ == "__main__" : 
       
    N = 4
   
    # Function Calling 
    print(NthTerm(N))

// C# program to find N-th
// term of the series:
using System;
 
class GFG
{
 
// function to return nth
// term of the series
static long NthTerm(long n)
{
    long x = (2 * n * n) % 1000000009;
    return (x - n + 1000000009) %
                    1000000009;
}
 
// Driver Code
public static void Main()
{
 
    // Taking n as 6
    long N = 4;
 
    // Printing the nth term
    Console.WriteLine(NthTerm(N));
}
}
 
// This code is contributed
// by inder_verma

<?php
// PHP program to find Nth
// term of the series
 
$mod = 1000000009;
 
// function to return nth
// term of the series
function NthTerm($n)
{
    global $mod;
    $x = (2 * $n * $n) % $mod;
    return ($x - $n + $mod) % $mod;
}
 
// Driver code
$N = 4;
 
// function call
echo NthTerm($N);
 
// This code is contributed
// by inder_verma
?>

<script>
 
// Javascript program to find N-th
// term of the series:
 
    // function to return nth term of the series
    function NthTerm(n) {
        var x = (2 * n * n) % 1000000009;
        return (x - n + 1000000009) % 1000000009;
    }
 
    // Driver Code
     
 
        // Taking n as 6
        var N = 4;
 
        // Printing the nth term
        document.write(NthTerm(N));
 
// This code contributed by gauravrajput1
 
</script>

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