Find the nth term of the given series

Given the first two terms of the series as 1 and 6 and all the elements of the series are 2 less than the mean of the number preceding and succeeding it. The task is to print the n^th term of the series. 
First few terms of the series are: 
 

1, 6, 15, 28, 45, 66, 91, ...

Examples: 
 

Input: N = 3 
Output: 15
Input: N = 1 
Output: 1 
 

Approach: The given series represents odd positioned numbers in the triangular number series. Since the n^th triangular number can easily be found by (n * (n + 1) / 2), so for finding the odd numbers we can replace n by (2 * n) - 1 as (2 * n) - 1 will always result in odd numbers i.e. the n^th number of the given series will be ((2 * n) - 1) * n.
Below is the implementation of the above approach: 
 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the nth term
// of the given series
int oddTriangularNumber(int N)
{
    return (N * ((2 * N) - 1));
}

// Driver code
int main()
{
    int N = 3;
    cout << oddTriangularNumber(N);

    return 0;
}

// Java implementation of the approach
class GFG
{

// Function to return the nth term
// of the given series
static int oddTriangularNumber(int N)
{
    return (N * ((2 * N) - 1));
}

// Driver code
public static void main(String[] args) 
{
    int N = 3;
    System.out.println(oddTriangularNumber(N));
}
}

// This code contributed by Rajput-Ji

# Python 3 implementation of the approach

# Function to return the nth term
# of the given series
def oddTriangularNumber(N):
    return (N * ((2 * N) - 1))

# Driver code
if __name__ == '__main__':
    N = 3
    print(oddTriangularNumber(N))

# This code is contributed by
# Surendra_Gangwar

// C# implementation of the approach 
using System;

class GFG 
{ 

    // Function to return the nth term 
    // of the given series 
    static int oddTriangularNumber(int N) 
    { 
        return (N * ((2 * N) - 1)); 
    } 
    
    // Driver code 
    public static void Main(String[] args) 
    { 
        int N = 3; 
        Console.WriteLine(oddTriangularNumber(N)); 
    } 
} 

/* This code contributed by PrinciRaj1992 */

<?php
// PHP implementation of the approach 

// Function to return the nth term 
// of the given series 
function oddTriangularNumber($N) 
{ 
    return ($N * ((2 * $N) - 1)); 
} 

    // Driver code 
    $N = 3; 
    echo oddTriangularNumber($N); 
    
    // This code is contributed by Ryuga

?>

<script>
// Javascript implementation of the approach

// Function to return the nth term
// of the given series
function oddTriangularNumber(N)
{
    return (N * ((2 * N) - 1));
}

// Driver code
let N = 3;
document.write(oddTriangularNumber(N));

// This code is contributed by subham348.
</script>

15

Time Complexity: O(1)

Auxiliary Space: O(1)