Sequences and Series: Diagram-Based Questions (1)

easy • 2 min read
Tags Sequences And Series

Question

A square has side 11 unit. A second square is inscribed by joining the midpoints of the first square’s sides. A third is inscribed in the second the same way, and so on indefinitely. Find the sum of the areas of all squares formed in this process.

Solution — Step by Step

The first square has side a1=1a_1 = 1. The second square’s side is the distance between adjacent midpoints — by Pythagoras on a half-square triangle, it’s (1/2)2+(1/2)2=12\sqrt{(1/2)^2 + (1/2)^2} = \tfrac{1}{\sqrt{2}}.

So the side ratio is 1/21/\sqrt{2} at each stage.

Areas: An=an2A_n = a_n^2. Since an+1/an=1/2a_{n+1}/a_n = 1/\sqrt{2}, An+1/An=1/2A_{n+1}/A_n = 1/2.

A1=1,A2=12,A3=14,…A_1 = 1, \quad A_2 = \tfrac{1}{2}, \quad A_3 = \tfrac{1}{4}, \quad \dots S=A11−r=11−1/2=2S = \frac{A_1}{1 - r} = \frac{1}{1 - 1/2} = 2

Sum is finite (= 2) because the ratio ∣r∣=1/2<1|r| = 1/2 < 1. The visual is consistent: each square is half the previous, the squares spiral inward toward the centre.

Total area = 2 square units.

Why This Works

When inscribing a square by midpoints, the new side is the diagonal of a small half-square — which equals the side divided by 2\sqrt{2}. Squaring gives an area ratio of exactly 12\tfrac{1}{2}.

The infinite-GP sum formula S∞=a/(1−r)S_\infty = a/(1-r) requires ∣r∣<1|r| < 1. Here r=1/2r = 1/2, well within bounds. The result is finite even though we add infinitely many positive terms — the key feature of a convergent geometric series.

Alternative Method

Direct partial sum: Sn=1+1/2+1/4+…+1/2n−1=2(1−1/2n)S_n = 1 + 1/2 + 1/4 + \ldots + 1/2^{n-1} = 2(1 - 1/2^n). As n→∞n \to \infty, 1/2n→01/2^n \to 0, so Sn→2S_n \to 2. Same answer through the partial-sum limit.

Common Mistake

Students sometimes compute the side ratio as 1/21/2 instead of 1/21/\sqrt{2} — confusing “half the side” with “midpoint construction.” The midpoints are joined, but the segment connecting them is the diagonal of the corner triangle, not a side. Always re-derive with Pythagoras when in doubt.

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