AS June 2024 Q5

EdexcelCurrent spec9 marksRecurrence Relations

Figure 1: three stages of a pattern; stage 1 is one large square, stage 2 is four smaller squares (top-left, centre, bottom-left and bottom-right corners, the top-right one removed), stage 3 repeats the process on each of those squares — Figure 1

Figure 1 shows the first three stages of a pattern that is created by a recursive process.

The process starts with a square and proceeds as follows

each square is replaced by 5 smaller squares each \(\dfrac{1}{9}\)th the size of the square being replaced
the 5 smaller squares are the ones in each corner and the one in the centre
once each of the squares has been replaced, the square immediately to the right and above the centre square of the pattern is then removed

Let \(u_n\) be the number of squares in the pattern in stage \(n\), where stage 1 is the original square.

(a) Explain why \(u_n\) satisfies the recurrence system\[u_1 = 1 \qquad u_{n+1} = 5u_n - 1 \qquad (n = 1, 2, 3, \ldots)\] (2)

(b) Solve this recurrence system. (5)

Given that the initial square has area 25

(c) determine the total area of all the squares in stage 8 of the pattern, giving your answer to 2 significant figures. (2)

Mark scheme (a)

Scheme	Marks	AO
In the first stage there is one square so \(u_1 = 1\) Each square from \(u_n\) to \(u_{n+1}\) is replaced by 5 smaller squares, so \(u_{n+1} = 5u_n\) But one of the squares is then removed, so \(u_{n+1} = 5u_n - 1\)	B1 B1	2.4 3.3
	(2)

Notes

B1: For explaining any two of the three aspects in the scheme.

B1: All three aspects explained.

Mark scheme (b)

Scheme	Marks	AO
AE is \(\lambda - 5 = 0 \Rightarrow \lambda = 5\)	M1	1.1b
So CF is \(w_n = A \times 5^n\)	A1	1.1b
PS try \(v_n = k \Rightarrow k = 5k - 1 \Rightarrow k = \ldots \Rightarrow u_n = \text{“}A \times 5^n\text{”} + \text{“}\tfrac{1}{4}\text{”}\)	M1	1.1b
\(u_1 = 1 \Rightarrow 1 = A \times 5^1 + \dfrac{1}{4} \Rightarrow A = \dfrac{3}{20}\)	M1	3.4
So \(u_n = \dfrac{3}{20} \times 5^n + \dfrac{1}{4}\) or \(u_n = \dfrac{3}{4} \times 5^{n-1} + \dfrac{1}{4}\) oe	A1	1.1b
	(5)

Notes

M1: Sets up and solves the auxiliary equation.

A1: Correct complementary part found.

M1: Selects correct form for particular solution and substitutes and combines result with their CF

M1: Uses the initial value to find the constant.

A1: Correct solution.

Mark scheme (c)

Scheme	Marks	AO
Each square in stage \(n\) has area \(\dfrac{25}{9^{n-1}}\) so total area is \(\dfrac{25}{9^7} \times u_8 = \dfrac{25}{9^7} \times \left(\dfrac{3}{4} \times 5^7 + \dfrac{1}{4}\right)\)	M1	3.1a
\(= 0.3062\ldots\) Accept awrt 0.31	A1	1.1b
	(2)
	(9 marks)

Notes

M1: Attempts a scale factor with their \(u_8\). Accept attempts at scaling by \(25 \times 3^{-k}\) or \(25 \times 9^{-k}\) where \(k\) is 7, 8 or 9.

A1: Correct answer.