Observe the Sierpiński triangle.

(A) How many black triangles are there in Stages 0 to 3?

(B) Predict the number at Stages 4 and 5.

(C) Find a rule for the number of black triangles at the $n^{\text{th}}$ stage.

(D) Suppose the area of the triangle (that is, the black region) in Stage 0 is 1 square unit. What is the area of the black region in Stages 1, 2 and 3? What will be the area of the black region in Stages 4 and 5? Find a rule for the area of the black region at the $n^{\text{th}}$ stage. What happens to this area as $n$, the number of stages, goes on increasing?                                                                                                                                                                                              [Page No. 189]

(A) Counting the black triangles, Stages 0, 1, 2 and 3 contain 1, 3, 9 and 27 black triangles respectively, because each black triangle is replaced by three smaller ones at the next stage.

(B) Continuing the $\times 3$ rule, Stage 4 has $27 \times 3 = 81$ and Stage 5 has $81 \times 3 = 243$ black triangles.

(C) The numbers 1, 3, 9, 27, 81, 243 are all powers of 3 ($3^0, 3^1, 3^2, \dots$), and the exponent matches the stage number. So, the number of black triangles at the $n$th stage is $3^n$.
(As a recursive rule: $t_1 = 1$ and $t_n = 3 \times t_{n-1}$.)

(D) At each stage the central piece is removed, leaving $\frac{3}{4}$ of the previous black area. So the area is multiplied by $\frac{3}{4}$ every stage:

So, the area at the $n$-th stage is $(\frac{3}{4})^n$ square units. Since $\frac{3}{4}$ is less than 1, multiplying by it again and again makes the area smaller and smaller, getting closer and closer to 0 as $n$ increases.
In other words, the number of black triangles grows very rapidly while the total black area shrinks towards zero.