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 = 3Output:15Input:N = 1Output: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**number of the given series will be

^{th}**((2 * n) – 1) * n**.

Below is the implementation of the above approach:

## C++

`// 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

`// 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` |

## Python3

`# 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#

`// 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` `// 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` `?>` |

## Javascript

`<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

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