# Tetracontagon Number

Given a number N, the task is to find Nth Tetracontagon number.

A Tetracontagon number is class of figurate number. It has 40 – sided polygon called tetracontagon. The N-th tetracontagon number count’s the 40 number of dots and all others dots are surrounding with a common sharing corner and make a pattern. The first few tetracontagonol numbers are 1, 40, 117, 232 …

Examples:

Input: N = 2
Output: 40
Explanation:
The second tetracontagonol number is 40.

Input: N = 3
Output: 117

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The N-th tetracontagon number is given by the formula:

• Nth term of s sided polygon = • Therefore Nth term of 40 sided polygon is Below is the implementation of the above approach:

## C++

 // C++ program for above approach  #include  using namespace std;     // Finding the nth tetracontagon Number  int tetracontagonNum(int n)  {      return (38 * n * n - 36 * n) / 2;  }     // Driver Code  int main()  {      int n = 3;      cout << "3rd tetracontagon Number is = "          << tetracontagonNum(n);         return 0;  }     // This code is contributed by Akanksha_Rai

## C

 // C program for above approach     #include  #include     // Finding the nth tetracontagon Number  int tetracontagonNum(int n)  {      return (38 * n * n - 36 * n) / 2;  }     // Driver program to test above function  int main()  {      int n = 3;      printf("3rd tetracontagon Number is = %d",             tetracontagonNum(n));         return 0;  }

## Java

 // Java program for above approach  import java.util.*;     class GFG {     // Finding the nth tetracontagon number  static int tetracontagonNum(int n)  {      return (38 * n * n - 36 * n) / 2;  }     // Driver code  public static void main(String[] args)  {      int n = 3;             System.out.println("3rd tetracontagon Number is = " +                                       tetracontagonNum(n));  }  }     // This code is contributed by offbeat

## Python3

 # Python3 program for above approach      # Finding the nth tetracontagon Number   def tetracontagonNum(n):          return (38 * n * n - 36 * n) // 2    # Driver Code  n = 3 print("3rd tetracontagon Number is = ",                      tetracontagonNum(n))      # This code is contributed by divyamohan123

## C#

 // C# program for above approach  using System;     class GFG {     // Finding the nth tetracontagon number  static int tetracontagonNum(int n)  {      return (38 * n * n - 36 * n) / 2;  }     // Driver code  public static void Main(string[] args)  {      int n = 3;             Console.Write("3rd tetracontagon Number is = " +                                  tetracontagonNum(n));  }  }     // This code is contributed by rutvik_56

Output:

3rd tetracontagon Number is = 117


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