Here’s a mathematical investigation based on the figure in the uploaded image:

(1) Introduction

The figure above represents a fractal-like triangular structure, where the number of black triangles increases progressively. At each stage $n$, smaller triangles are formed within the original triangle. The purpose of this investigation is to determine and explore the relationship between the stage $n$ and the number of black triangles present.

(2) Problem to be Explored

How many black triangles exist in Figure $n$, and can a general formula be derived to calculate this number?

(3) Conjecture 1

The number of black triangles follows a geometric progression. Specifically, the total number of black triangles at stage $n$ appears to be $3^{n-1}$, where $n$ is the figure’s stage number.

(4) Verification

We verify the conjecture by analyzing the figures:

• Stage 1 ($n = 1$): 1 black triangle.
• Stage 2 ($n = 2$): 3 black triangles.
• Stage 3 ($n = 3$): 9 black triangles.
• Stage 4 ($n = 4$): 27 black triangles.

The number of black triangles follows the progression $3^{n-1}$.

(5) Prove

The fractal nature of the figure divides the triangle into smaller equal sub-triangles at each stage:

1. Initially, $1$ black triangle occupies the entire space.
2. At stage $n$, each black triangle from stage $n-1$ is divided into 3 smaller triangles.
3. Thus, at stage $n$, the total number of black triangles is given by: $T_n = 3 \times T_{n-1}$ Using $T_1 = 1$ as the base case: $T_n = 3^{n-1}$

(6) Observation

The number of black triangles increases exponentially with each stage. The fractal structure demonstrates self-similarity and recursive division.

(7) Conjecture 2

The total number of all triangles (black and white) at any stage $n$ can be calculated as: $S_n = \frac{3^n - 1}{2}$

(8) Verification

• Stage 1 ($n = 1$): $S_1 = \frac{3^1 - 1}{2} = 1$.
• Stage 2 ($n = 2$): $S_2 = \frac{3^2 - 1}{2} = 4$.
• Stage 3 ($n = 3$): $S_3 = \frac{3^3 - 1}{2} = 13$.
• Stage 4 ($n = 4$): $S_4 = \frac{3^4 - 1}{2} = 40$.

The conjecture holds true for the total number of triangles.

(9) Prove

The number of total triangles at stage $n$ follows a recursive pattern:

1. At stage $n = 1$, there is 1 triangle.
2. At each subsequent stage, every triangle from the previous stage is divided into 3 sub-triangles: $S_n = S_{n-1} + 3^{n-1}$ Using summation: $S_n = \sum_{k=1}^{n} 3^{k-1} = \frac{3^n - 1}{2}$

(10) Observation

The structure illustrates geometric growth and symmetry. Black triangles represent part of a fractal set, while the total triangles (black and white) combine recursively to demonstrate exponential patterns.

Would you like further details or graphical representations of these findings?

Related Questions:

1. How does this pattern relate to Sierpiński triangles?
2. What is the total area of black triangles as $n \to \infty$?
3. What is the ratio of black to white triangles at each stage?
4. How does the perimeter change as $n$ increases?
5. Can the formula $T_n = 3^{n-1}$ be generalized for other fractal patterns?

Tip:

Explore fractal geometry for more patterns that combine recursion and self-similarity!