A2 June 2023 Q7

EdexcelCurrent spec8 marksRecurrence Relations

7. Martina decides to open a bank account to help her to save for a holiday. Each month she puts £\(k\) into the account and allows herself to spend one quarter of what was in the account at the end of the previous month.

Let \(u_n\) (where \(n \geqslant 1\)) represent the amount in the account at the end of month \(n\).

Martina has £\(k\) in the account at the end of the first month.

(a) By setting up a first order recurrence relation for \(u_{n+1}\) in terms of \(u_n\), determine an expression for \(u_n\) in terms of \(n\) and \(k\). (6)

At the end of the 8th month, Martina needs to have at least £1750 in the account to pay for her holiday.

(b) Determine, to the nearest penny, the minimum amount of money that Martina should put into the account each month. (2)

Mark scheme (a)

Scheme	Marks	AO
\(u_{n+1} = 0.75u_n + k\)	B1	3.3
(aux equation \(m - 0.75 = 0 \Rightarrow\)) complementary function is \(A(0.75)^n\)	B1	1.1b
Consider a trial solution of the form \(u_n = \lambda\) so \(\lambda - 0.75\lambda = k\)	M1	1.1b
General solution is \(u_n = A(0.75)^n + 4k\)	A1	1.1b
\(u_1 = k \therefore 0.75A + 4k = k \Rightarrow A = \ldots\)	M1	3.4
\(u_n = 4k\left(1 - (0.75)^n\right)\)	A1	2.2a
	(6)

Notes

B1: CAO

M1: substituting their trial solution into the recurrence relation (which must be of the form \(u_{n+1} = \alpha u_n + k\) where \(\alpha = 0.75\) or \(\pm 0.25\)) in an attempt to find their \(\lambda\)

A1: CAO for the general solution (may be implied by subsequent working)

M1: using the conditions in the model and their expression of the form \(A(\alpha)^n + Bk\) where \(\alpha\) is 0.75 or \(\pm 0.25\) to calculate \(A\) (which if correct is \(-4k\))

A1: CAO for the particular solution (may be an expression without \(u_n =\) accept any equivalent form e.g. \(u_n = 4k - 3k(0.75)^{n-1}\))

Mark scheme (b)

Scheme	Marks	AO
\(4k\left(1 - 0.75^8\right) = 1750 \Rightarrow k = \ldots\)	dM1	3.4
\(k = 486.1721068\ldots \Rightarrow k = 486.18\)	A1	2.2a
	(2)
	(8 marks)

Notes

dM1: Applying \(n = 8\) to their general solution of the form \(A(\alpha)^n + Bk\) where \(\alpha\) is 0.75 or \(\pm 0.25\) equal to 1750 (dependent on both M marks in (a))

A1: CAO – must be 486.18