MRSM Trial 2025 Add Maths Paper 2, Question 6Progressions
Solved on video in English · Form 4, Progressions · 4:47 video
The question
Diagram 4 shows the steps to produce triangles in Sierpinski's model. For every step, each triangle is divided into four congruent triangles and the middle triangle is removed. This process is repeated in each subsequent step.
(a) The number of removed triangles in the n-th step, Tₙ, forms a progression. Show that Tₙ = 3ⁿ/3.
Diagram 4 shows the steps to produce triangles in Sierpinski's model (each triangle divided into four congruent triangles, the middle one removed, repeated each step).
(b) Find the sum of triangles that have not been removed for the first 8 steps.
(c) Given that the area of the triangle removed in the first step is 12 unit², find the sum to infinity of the areas of the triangles that have been removed.
4:47Part (a)
Step 1
Count removed triangles: Step 1 removes 1, Step 2 removes 3, Step 3 removes 9 — a geometric progression with first term a = 1 and common ratio r = 3.
Why this step
Each count is 3 times the one before, so it's a geometric progression. Spotting that tells us which formulas we can use.
1 more step, with the reason for it, in the full solution
Final answer
Shown: Tₙ = 3ⁿ⁻¹ = 3ⁿ/3
Where students lose marks
Watch out: we want one term, so use Tₙ = arⁿ⁻¹, not the sum formula, and keep the exponent n − 1.
Part (b)
3 more steps, each with the reason for it, in the full solution
Final answer
9840
Where students lose marks
Watch out: sum the triangles left (a = 3), not the removed ones (a = 1). That wrongly gives 3280.
Part (c)
3 more steps, each with the reason for it, in the full solution
Final answer
48 unit²
Where students lose marks
Watch out: the ratio is 3 × ¼ = 3/4, not 1/4 or 3, and use a/(1 − r), not the sum-to-n formula.
Watch every step of this question solved on video (4:47), with the reason behind each step and a box that checks your own answer.
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