Diagram 1 shows an equilateral triangle with sides of length 1 unit. An inverted white equilateral triangle with sides of length $\frac{1}{2}$ unit is removed to form Diagram 2. Three inverted white equilateral triangles with sides of length $\frac{1}{4}$ unit is then removed from Diagram 2 to form Diagram 3. The process continues in this way.

1. Identify the sequence for the number of black triangles in each diagram

Diagram 1: $1$ black triangle
Diagram 2: $3$ black triangles
Diagram 3: $9$ black triangles
Diagram 4: $27$ black triangles
Diagram 5: $81$ black triangles

The sequence is $1, 3, 9, 27, 81, \dots$

2. Deduce the general formula for the $n$-th term

Notice that each term is a power of $3$:
Diagram 1: $3^0$
Diagram 2: $3^1$
Diagram 3: $3^2$

For Diagram $n$, the power of $3$ is $n-1$.
Number of black triangles $= 3^{n-1}$

1. Determine the area of black triangles in Diagram 2

Area of Diagram 1 = $a$
In Diagram 2, an inverted triangle with side length $\frac{1}{2}$ is removed. The area of this removed triangle is $\left(\frac{1}{2}\right)^2 = \frac{1}{4}$ of the original area.
Area of black triangles in Diagram 2 = $a - \frac{1}{4}a = \frac{3}{4}a$

2. Generalise the pattern to Diagram $n$

In each subsequent diagram, the area of black triangles is $\frac{3}{4}$ of the previous diagram's area.
Area in Diagram 1 = $a$
Area in Diagram 2 = $a\left(\frac{3}{4}\right)^1$
Area in Diagram 3 = $a\left(\frac{3}{4}\right)^2$
Area in Diagram $n$ = $a\left(\frac{3}{4}\right)^{n-1}$