AS June 2018 Q3

EdexcelCurrent spec10 marksRecurrence Relations

3. A tree at the bottom of a garden needs to be reduced in height. The tree is known to increase in height by 15 centimetres each year.

On the first day of every year, the height is measured and the tree is immediately trimmed by 3% of this height.

When the tree is measured, before trimming on the first day of year 1, the height is 6 metres.

Let \(H_n\) be the height of the tree immediately before trimming on the first day of year \(n\).

(a) Explain, in the context of the problem, why the height of the tree may be modelled by the recurrence relation\[H_{n+1} = 0.97H_n + 0.15, \quad H_1 = 6, \quad n \in \mathbb{Z}^+\] (3)

(b) Prove by induction that \(H_n = 0.97^{n-1} + 5, \quad n \geqslant 1\) (4)

(d) By what fixed percentage should the tree be trimmed each year if the height of the tree immediately before trimming is to be 4 metres in the long term? (2)

Mark scheme (a)

Scheme	Marks	AO
\(H_n\) is the measured height at the start of year \(n\) and this is decreased by 3% at the start of year \(n\), so is multiplied by 97% = 0.97 to give \(0.97H_n\) as the new height due to trimming	B1	3.3
0.15 is added to \(0.97H_n\) as 0.15 is 15 cm in m and this is how much the tree grows in a year.	B1	3.4
And \(H_1 = 6\) is the height of the tree at the start of year 1 before trimming	B1	1.1b
	(3)

Notes

B1: Need to see 3% decrease linked to scale factor of 0.97

B1: Need to see that adding 0.15 corresponds to the yearly growth in metres. There must be some reference to the units for this mark.

B1: An explanation that \(H_1\) is the first term (the starting height) and this is 6m

Mark scheme (b)

Scheme	Marks	AO
\(n = 1 \Rightarrow H_1 = (0.97)^{1-1} + 5 = 6\) So true for \(n = 1\)	B1	2.1
Assume true for \(n = k\) so \(H_k = (0.97)^{k-1} + 5\) so \(H_{k+1} = 0.97\left((0.97)^{k-1} + 5\right) + 0.15\)	M1	2.4
so \(H_{k+1} = (0.97)^k + 4.85 + 0.15 = (0.97)^k + 5\)	A1	1.1b
If true for \(n = k\) then true for \(n = k + 1\), true for \(n = 1\) so true for all (positive integers) \(n\) (Allow “for all values”)	B1	2.2a
	(4)

Notes

B1: Begins proof by induction by considering \(n = 1\) and obtains \(H_1 = 6\)

M1: Assumes true for \(n = k\) and uses iterative formula to consider \(n = k + 1\)

A1: Reaches \((0.97)^k + 5\) with no errors

B1: Correct conclusion. This mark is dependent on all previous marks apart from the first B mark. It is gained by conveying the ideas of all four underlined points either at the end of their solution or as a narrative in their solution.

Scheme	Marks	AO
The height will approach 5m	B1	1.1b
	(1)

Mark scheme (d)

Scheme	Marks	AO
Require \(4 = 4x + 0.15\)	M1	3.1b
\(x = 0.9625\) so 3.75%	A1	1.1b
	(2)
	(10 marks)

Notes

M1: Uses the model to adopt a correct strategy to find the required percentage

A1: Interprets their answer correctly in terms of the original context