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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 nth 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:
 

 

Approach: The given series represents odd positioned numbers in the triangular number series. Since the nth 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 nth 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>

Output: 
15

 

Time Complexity: O(1)

Auxiliary Space: O(1)


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