Count the total number of triangles after Nth operation

Given an equilateral triangle, the task is to compute the total number of triangles after performing the following operation N times.
For every operation, the uncolored triangles are taken and divided into 4 equal equilateral triangles. Every inverted triangle formed is colored. Refer to the below figure for more details.

For N=1 the triangle formed is:

For N=2 the triangle formed is:



Examples:

Input :N = 10
Output : 118097

Input : N = 2
Output : 17

Approach:

  • At every operation, 3 uncolored triangles, 1 colored triangle and the triangle itself is formed
  • On writing the above statement mathematically; count of triangles at Nth move = 3 * count of triangles at (N-1)th move + 2
  • Therefore, initializing a variable curr = 1 and tri_count = 0
  • Next, a loop is iterated from 1 to N
  • For every iteration, the operation mentioned above is performed.
  • Finally, the tri_count is returned
  • Below is the implementation of the above approach:

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    #include <bits/stdc++.h>
    using namespace std;
    // function to return the
    // total no.of Triangles
    int CountTriangles(int n)
    {
        int curr = 1;
        int Tri_count = 0;
        for (int i = 1; i <= n; i++) {
            // For every subtriangle formed
            // there are possibilities of
            // generating (curr*3)+2
      
            Tri_count = (curr * 3) + 2;
            // Changing the curr value to Tri_count
            curr = Tri_count;
        }
        return Tri_count;
    }
      
    // driver code
    int main()
    {
        int n = 10;
        cout << CountTriangles(n);
        return 0;
    }
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    class Gfg {
        // Method to return the
        // total no.of Triangles
        public static int CountTriangles(int n)
        {
            int curr = 1;
            int Tri_count = 0;
            for (int i = 1; i <= n; i++) {
                // For every subtriangle formed
                // there are possibilities of
                // generating (curr*3)+2
      
                Tri_count = (curr * 3) + 2;
                // Changing the curr value to Tri_count
                curr = Tri_count;
            }
            return Tri_count;
        }
      
        // driver code
        public static void main(String[] args)
        {
            int n = 10;
            System.out.println(CountTriangles(n));
        }
    }
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    # Function to return the 
    # total no.of Triangles
    def countTriangles(n):
          
        curr = 1
        Tri_count = 0
        for i in range(1, n + 1):
                  
            # For every subtriangle formed
            # there are possibilities of 
            # generating (curr * 3)+2
            Tri_count = (curr * 3) + 2
            # Changing the curr value to Tri_count
            curr = Tri_count
        return Tri_count
          
    n = 10
    print(countTriangles(n))
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    using System;
      
    class Gfg 
    {
        // Method to return the
        // total no.of Triangles
        public static int CountTriangles(int n)
        {
            int curr = 1;
            int Tri_count = 0;
            for (int i = 1; i <= n; i++) 
            {
                // For every subtriangle formed
                // there are possibilities of
                // generating (curr*3)+2
                Tri_count = (curr * 3) + 2;
                  
                // Changing the curr value to Tri_count
                curr = Tri_count;
            }
            return Tri_count;
        }
      
        // Driver code
        public static void Main(String[] args)
        {
            int n = 10;
            Console.WriteLine(CountTriangles(n));
        }
    }
      
    // This code is contributed by 29AjayKumar
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    Output:
    118097
    

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