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

1. Identify the pattern in the number of black triangles

Diagram 1 has $1$ black triangle.

Diagram 2 has $3$ black triangles.

Diagram 3 has $9$ black triangles.

So the number of black triangles follows
$1,\ 3,\ 9,\ldots$
with each term multiplied by $3$.

2. Write the general expression for Diagram $n$

Since Diagram 1 corresponds to $3^0$, Diagram 2 to $3^1$, and Diagram 3 to $3^2$, Diagram $n$ corresponds to
$3^{n-1}.$

1. Find the scale factor for total black area from one diagram to the next

From one diagram to the next, each black triangle is replaced by $3$ smaller black triangles.

Each smaller triangle has side length half of the previous one, so its area is
$\left(\frac{1}{2}\right)^2=\frac{1}{4}$
of the previous black triangle.

So the total black area is multiplied by
$3\times \frac{1}{4}=\frac{3}{4}.$

2. Write the general expression for the total black area in Diagram $n$

Diagram 1 has black area $a$.

There are $n-1$ stages from Diagram 1 to Diagram $n$, so
$\text{Area in Diagram }n=a\left(\frac{3}{4}\right)^{n-1}.$