Follow the rule, one stage at a time. The paragraph tells us exactly what happens at each step: every red (shaded) square is cut by trisection into \(\displaystyle 3\times 3=9\) equal small squares, the middle one is thrown away, and the remaining \(\displaystyle 8\) stay red. So going from one stage to the next,
every red square is replaced by \(\displaystyle 8\) red squares, each with side one-third as long.
(i) Red squares in Stages $\displaystyle 0$ to 3.Stage $\displaystyle 0$: the whole sheet is one red square, so \(\displaystyle 1\) red square.
Stage $\displaystyle 1$: the one square becomes \(\displaystyle 8\), so \(\displaystyle 8\) red squares.
Stage $\displaystyle 2$: each of those \(\displaystyle 8\) becomes \(\displaystyle 8\), so \(\displaystyle 8\times 8=64\).
Stage $\displaystyle 3$: each of those \(\displaystyle 64\) becomes \(\displaystyle 8\), so \(\displaystyle 8\times 64=512\).
(ii) Predicting Stages $\displaystyle 4$ and 5. The rule "multiply by $\displaystyle 8$" does not change, so
\[\text{Stage }4:\ 8\times 512=4096,\qquad \text{Stage }5:\ 8\times 4096=32768.\]
(iii) A rule for the \(\displaystyle n^{\text{th}}\) stage. The counts \(\displaystyle 1,8,64,512,\ldots\) form a GP with first term \(\displaystyle 1\) (at Stage $\displaystyle 0$) and common ratio \(\displaystyle 8\). Writing \(\displaystyle S_n\) for the number of red squares at Stage \(\displaystyle n\),
\[\textbf{Explicit: }\ S_n=8^{\,n},\qquad\qquad \textbf{Recursive: }\ S_0=1,\ \ S_n=8\,S_{n-1}\ \ (n\ge 1).\]
The explicit formula lets you jump straight to any stage; the recursive one says in symbols what the construction actually does.
(iv) Area of the red region. Take the Stage $\displaystyle 0$ square to have area \(\displaystyle 1\) square unit, so its side is \(\displaystyle 1\) unit. Trisecting the side gives small squares of side \(\displaystyle \tfrac13\), and area
\[\left(\tfrac13\right)^{2}=\tfrac19\ \text{of the square they came from}.\]
Eight of the nine pieces are kept, so at every stage
\[\text{red area}\ \longrightarrow\ 8\times\frac{1}{9}\times(\text{red area})=\frac{8}{9}\times(\text{red area}).\]
Hence, with \(\displaystyle A_n\) the red area at Stage \(\displaystyle n\):
\[A_1=\frac{8}{9},\qquad A_2=\frac{64}{81},\qquad A_3=\frac{512}{729}.\]
\[A_4=\frac{4096}{6561},\qquad A_5=\frac{32768}{59049}.\]
The same numbers come out if you count directly: at Stage \(\displaystyle n\) there are \(\displaystyle 8^{n}\) red squares, each of side \(\displaystyle \left(\tfrac13\right)^{n}\), so the red area is \(\displaystyle 8^{n}\cdot\left(\tfrac{1}{3^{n}}\right)^{2}=\dfrac{8^{n}}{9^{n}}\). That is a useful second check.
\[\textbf{Explicit: }\ A_n=\left(\frac{8}{9}\right)^{n},\qquad\qquad \textbf{Recursive: }\ A_0=1,\ \ A_n=\frac{8}{9}A_{n-1}\ \ (n\ge 1).\]
| Stage \(\displaystyle n\) | Red squares \(\displaystyle S_n\) | Side of each square | Red area \(\displaystyle A_n\) |
| $\displaystyle 0$ | $\displaystyle 1$ | $\displaystyle 1$ | $\displaystyle 1$ |
| $\displaystyle 1$ | $\displaystyle 8$ | \(\displaystyle 1/3\) | \(\displaystyle 8/9\approx 0.889\) |
| $\displaystyle 2$ | $\displaystyle 64$ | \(\displaystyle 1/9\) | \(\displaystyle 64/81\approx 0.790\) |
| $\displaystyle 3$ | $\displaystyle 512$ | \(\displaystyle 1/27\) | \(\displaystyle 512/729\approx 0.702\) |
| $\displaystyle 4$ | $\displaystyle 4096$ | \(\displaystyle 1/81\) | \(\displaystyle 4096/6561\approx 0.624\) |
| $\displaystyle 5$ | $\displaystyle 32768$ | \(\displaystyle 1/243\) | \(\displaystyle 32768/59049\approx 0.555\) |
What happens as \(\displaystyle n\) increases? Each stage multiplies the area by \(\displaystyle \tfrac89\), and \(\displaystyle \tfrac89<1\), so the red area gets smaller every single time and never stops shrinking. Multiplying by \(\displaystyle \tfrac89\) again and again drives \(\displaystyle \left(\tfrac89\right)^{n}\) as close to \(\displaystyle 0\) as we like — for instance \(\displaystyle \left(\tfrac89\right)^{50}\approx 0.003\). So the red area tends to \(\displaystyle 0\): in the "limit" the carpet has
more and more squares but
less and less area. (Note it never actually equals \(\displaystyle 0\) at any finite stage.)
Answers: (i) $\displaystyle 1$, $\displaystyle 8$, $\displaystyle 64$, 512. (ii) $\displaystyle 4096$ and 32768. (iii) \(\displaystyle S_n=8^{n}\); \(\displaystyle S_0=1,\ S_n=8S_{n-1}\). (iv) \(\displaystyle \tfrac{8}{9},\ \tfrac{64}{81},\ \tfrac{512}{729}\), then \(\displaystyle \tfrac{4096}{6561}\) and \(\displaystyle \tfrac{32768}{59049}\); \(\displaystyle A_n=\left(\tfrac{8}{9}\right)^{n}\) with \(\displaystyle A_0=1,\ A_n=\tfrac{8}{9}A_{n-1}\); and the red area shrinks towards $\displaystyle 0$ as \(\displaystyle n\) grows. 