Stand Up & Math · Teacher Answers
Forever Squared
Not for distribution · Geometric sequences & series · Full worked solutions
"A designer starts with a single square. At every stage, new squares are attached to every free edge — each one half the side length of the square it touches. The pattern keeps growing. Your job today: figure out how much area it covers — and what happens if it grows forever."
Display the fractal diagram on student devices when Phase 1 unlocks. Students work entirely on their whiteboard.
Geometric sequences · constant ratio · nth term
| Stage \(n\) | New squares | Side length | Area each | New area added |
|---|---|---|---|---|
| 0 | 1 | 1 | 1 | 1 |
| 1 | 4 | \(\tfrac{1}{2}\) | \(\tfrac{1}{4}\) | 1 |
| 2 | 12 | \(\tfrac{1}{4}\) | \(\tfrac{1}{16}\) | \(\tfrac{3}{4}\) |
| 3 | 36 | \(\tfrac{1}{8}\) | \(\tfrac{1}{64}\) | \(\tfrac{9}{16}\) |
| 4 | 108 | \(\tfrac{1}{16}\) | \(\tfrac{1}{256}\) | \(\tfrac{27}{64}\) |
Stage 0 → Stage 1 also produces new area = 1. This means the "new area" column reads 1, 1, ¾, 9/16 … — the geometric behaviour with ratio ¾ only starts from Stage 1. When students spot this, encourage them: the geometric sequence is the series of new areas from Stage 1 onwards: \(1,\, \tfrac{3}{4},\, \tfrac{9}{16},\,\ldots\) with \(u_1 = 1,\ r = \tfrac{3}{4}\).
From Stage 1 onwards, each term of the new area column is multiplied by \(\dfrac{3}{4}\).
Each new stage multiplies the count of squares by 3 (each square has 3 free edges) and multiplies the area of each square by \(\tfrac{1}{4}\) (side halves). So the total new area ratio = \(3 \times \tfrac{1}{4} = \tfrac{3}{4}\).
Treating Stage 1 as the first term: \(a_n = \left(\dfrac{3}{4}\right)^{n-1}\) for \(n \geq 1\), where \(a_n\) is the new area added at stage \(n\).
Alternatively, if students write \(a_n\) for the new area at stage \(n\) (starting from \(n=1\)):
Verify: table completed correctly, \(r = 3/4\) identified with justification, and a working general expression for \(a_n\). The exact form of notation may vary — accept any equivalent expression.
Geometric series · deriving \(S_n\) · telescoping cancellation
Total area = Stage 0 area + sum of all new areas added from Stage 1.
Let \(S\) = sum of the geometric series from Stage 1 (new areas only, Stage 0 treated separately):
Total area after 10 stages = \(1 + S \approx 4.717\).
The key insight is that subtracting \(rS\) from \(S\) cancels all interior terms. Only the first term of \(S\) and the last term of \(rS\) survive. Ask groups to circle those two surviving terms before simplifying.
Sum of \(n\) new-area terms (Stages 1 through \(n\)):
Total area including Stage 0:
Verify: the telescoping subtraction written out in full, the formula \(S_n = \frac{u_1(1-r^n)}{1-r}\) derived (not recalled), and the total area formula including Stage 0. Numerical answer for 10 stages ≈ 4.717.
Infinite geometric series · \(S_\infty\) · convergence
The total clearly approaches 5 without ever reaching it.
As \(n \to \infty\): since \(|r| = \tfrac{3}{4} < 1\), we have \(r^n \to 0\).
Total area including Stage 0: \(1 + 4 = \mathbf{5}\).
Make sure groups articulate why \(r^n \to 0\): because \(r\) is a fraction less than 1, repeated multiplication makes it smaller and smaller. This is not obvious — ask them to compute \((3/4)^{10}\), \((3/4)^{50}\) and watch it shrink.
The fractal, grown forever, covers exactly 5 square units of area — starting from a single unit square.
Whether students find this surprising depends on their intuition. The key idea to surface: infinitely many terms can sum to a finite number, as long as the terms shrink fast enough.
Verify: \(S_\infty = 4\) written for the geometric series, total area = 5 stated clearly, and the condition \(|r| < 1\) linked to why the sum converges. Students should be able to explain in words why the sum doesn't keep growing forever.
Divergence · \(|r| \geq 1\) · the paradox
New squares added per stage: 1, 4, 12, 36, 108, …
From Stage 1: 4, 12, 36, 108 → ratio = 3 each time. Yes, this is geometric with \(u_1 = 4,\ r = 3\).
Total squares after \(n\) stages: \(1 + 4 + 12 + 36 + \cdots\)
Since \(r = 3 > 1\), the condition \(|r| < 1\) is not satisfied. As \(n \to \infty\), \(r^n \to \infty\), so the sum diverges — it grows without bound.
Infinitely many squares, yet total area = 5. This is not a contradiction because the squares become vanishingly small — each generation's squares are \(\tfrac{1}{4}\) the area of the previous generation's. The count grows (×3), but the areas shrink (×\(\tfrac{1}{4}\)), and the shrinking wins: net factor \(\tfrac{3}{4} < 1\).
A good sentence to draw out: "You can have infinitely many objects and still fit them in a finite space, as long as they get small enough fast enough."
SL minimum: Task 4a ratio = 3 identified, Task 4b divergence explained with the condition \(|r| \geq 1\), and Task 4c a written sentence on the board explaining the paradox.
HL: additionally, the HL extension solved correctly (see below).
Convergence condition: \(S_\infty\) exists only when \(|r| < 1\), i.e. \(-1 < r < 1\).
Find \(r\) such that total area = 3:
Check: \(|1/2| < 1\) ✓
r = 1/2 gives total area = 3
Select 2–3 groups · 10–12 min debrief