AS June 2019 Q5

EdexcelCurrent spec11 marksRecurrence Relations

5. On Jim’s 11th birthday his parents invest £1000 for him in a savings account.

The account earns 2% interest each year.

On each subsequent birthday, Jim’s parents add another £500 to this savings account.

Let \(U_n\) be the amount of money that Jim has in his savings account \(n\) years after his 11th birthday, once the interest for the previous year has been paid and the £500 has been added.

(a) Explain, in the context of the problem, why the amount of money that Jim has in his savings account can be modelled by the recurrence relation of the form\[U_n = 1.02U_{n-1} + 500 \qquad\qquad U_0 = 1000 \qquad n \in \mathbb{Z}^+\] (3)

(b) State an assumption that must be made for this model to be valid. (1)

Jim hopes to be able to buy a car on his 18th birthday.

(d) Use the answer to part (c) to find out whether Jim will have enough money in his savings account to buy a car that costs £4 500 (2)

Mark scheme (a)

Scheme	Marks	AO
\(U_{n-1}\) is the amount in the saving account \(n - 1\) years after Jim’s 11^th birthday. This is increased by 2% each year, so is multiplied by 1.02 to give \(1.02U_{n-1}\)	B1	3.3
Jim’s parents invest £500 for each subsequent birthday so 500 is added	B1	3.4
\(U_0 = 1000\) as this is the amount invested on Jim’s 11^th birthday	B1	1.1b
	(3)

Notes

B1: Need to explain that 2% interest rate linked to multiplication by scale factor 1.02

B1: Need to explain that 500 is added due to receiving £500 each year

B1: Needs to explain that \(U_0 = 1000\) is the initial amount invested

Mark scheme (b)

Scheme	Marks	AO
To use this model, one of, for example The interest rate stays the same each year Jim does not withdraw any money from the savings account Jim only saves the birthday money +£500 in this saving account, he does not invest any other money.	B1	3.5b
	(1)

Notes

B1: See main scheme

Mark scheme (c)

Scheme	Marks	AO
A complete method to solve the recurrence relation using \(U_n = \text{CF} + \text{PS} = c(1.02)^n + \lambda\)	M1	3.1a
\(\text{PS} = \lambda \quad \Rightarrow \lambda = 1.02\lambda + 500\) leading to \(\lambda = \ldots\)	M1	1.1b
\(\lambda = -25\,000\)	A1	1.1b
Uses \(U_0 = 1000\) and their value for \(\lambda\) to find the value of \(1000 = c(1.02)^0 - 25\,000\) \(c = \ldots(26\,000)\)	M1	1.1b
\(U_n = 26\,000(1.02)^n - 25\,000 \qquad (n \geqslant 0)\)	A1	1.1b
	(5)

Notes

M1: A complete method to solve the recurrence relation using \(U_n = \text{CF} + \text{PS} = c(1.02)^n + \lambda\)

M1: Uses \(\text{PS} = \lambda \quad \Rightarrow \lambda = 1.02\lambda + 500\) to find a value for \(\lambda\)

A1: \(\lambda = -25\,000\)

M1: Uses \(U_0\) and their value for \(\lambda\) to find a value of \(c\)

A1: Fully correctly defined sequence \(U_n = 26000(1.02)^n - 25\,000, \quad (n \geqslant 0)\)

Alternative 1

Scheme	Marks	AO
Realises that \(U_n =\) term of a GP + sum of a GP both with \(r = 1.02\)	M1	3.1a
Sum of a GP \(= \dfrac{500(1 - 1.02^n)}{1 - 1.02}\) or \(\dfrac{500(1.02^n - 1)}{1.02 - 1}\)	M1 A1	1.1b 1.1b
Term of a GP \(= 1000(1.02)^n\) or \(1000(1.02)^{n-1}\)	M1	1.1b
\(U_n = 1000(1.02)^n - 25\,000(1 - 1.02^n)\) or \(U_n = 1000(1.02)^n + 25\,000(1.02^n - 1)\)	A1	1.1b
	(5)

M1: A correct form for \(U_n\) term of a GP + Sum of a GP both with \(r = 1.02\)

M1: For the sum of a GP with \(a = 500\), \(r = 1.02\) and uses \(n\) or \(n - 1\)

A1: Correct the sum of a GP with \(a = 500\), \(r = 1.02\) and \(n\)

M1: For the term of a GP with \(a = 1000\), \(r = 1.02\) and uses \(n\) or \(n - 1\)

A1: Fully correctly defined sequence \(U_n\)

Mark scheme (d)

Scheme	Marks	AO
Uses \(U_n = 26\,000(1.02)^n - 25\,000\), with either \(n = 7\) or 8	M1	3.4
\(U_7 = 4865.83 \gt 4500\) therefore, Jim will have enough money in his savings account to buy a car costing £ 4500.	A1ft	2.2a
	(2)
	(11 marks)

Notes

M1: Uses their \(U_n\) with either \(n = 7\) or 8

A1ft: Finds \(U_7\) compares with 4 500 and comes to an appropriate conclusion. Follow through on their value of \(U_7\)