AS October 2020 Q4

EdexcelCurrent spec10 marksRecurrence Relations

4. Sam borrows £10 000 from a bank to pay for an extension to his house.
The bank charges 5% annual interest on the portion of the loan yet to be repaid.
Immediately after the interest has been added at the end of each year and before the start of the next year, Sam pays the bank a fixed amount, £\(F\).

Given that £\(A_n\) (where \(A_n \geqslant 0\)) is the amount owed at the start of year \(n\),

(a) write down an expression for \(A_{n+1}\) in terms of \(A_n\) and \(F\), (1)

(b) prove, by induction that, for \(n \geqslant 1\)\[A_n = (10\,000 - 20F)1.05^{n-1} + 20F\] (5)

Scheme	Marks	AO
\(A_{n+1} = 1.05A_n - F\)	B1	3.3
	(1)

Mark scheme (b)

Scheme	Marks	AO
\(n = 1 \Rightarrow A_1 = (10\,000 - 20F)1.05^{1-1} + 20F\) \(n = 1 \Rightarrow A_1 = 10\,000 - 20F + 20F = 10\,000\) So true for \(n = 1\)	B1	2.1
Assume true for \(n = k\) so that \(A_k = (10\,000 - 20F)1.05^{k-1} + 20F\) \(A_{k+1} = \ldots\)	M1	2.4
\(A_{k+1} = 1.05\left((10\,000 - 20F)1.05^{k-1} + 20F\right) - F\)	A1ft	1.1b
\(A_{k+1} = (10\,000 - 20F)1.05^k + 21F - F\) \(= (10\,000 - 20F)1.05^k + 20F\)	A1	1.1b
So the result holds for \(n = k + 1\) and so \(A_n = (10\,000 - 20F)1.05^{n-1} + 20F\) is true for all \(n \geqslant 1\)	A1	2.2a
	(5)

Notes

B1: Demonstrates that the result is true for \(n = 1\)

M1: Makes a statement that assumes the result is true for some value of \(n\)

A1ft: Correct expression for \(A_{k+1}\)

A1: Reaches the correct statement for \(k + 1\) in the required form

A1: Completes the inductive argument

Mark scheme (c)

Scheme	Marks	AO
\((10\,000 - 20F)1.05^{16-1} + 20F \leqslant 0\)	M1	3.1b
\(10\,000 \times 1.05^{15} \leqslant 20F(1.05^{15} - 1) \Rightarrow F \geqslant \ldots\)	M1	2.1
\(F \geqslant \dfrac{10\,000 \times 1.05^{15}}{20(1.05^{15} - 1)}\)	A1	1.1b
So the smallest value of \(F\) is £963.43 or So the smallest value of \(F\) is £964	A1	3.2a
	(4)
	(10 marks)

Notes

M1: Identifies a correct strategy using \(n = 16\) to obtain an equation in \(F\)

M1: Proceeds to obtain a value for the minimum value of \(F\)

A1: Correct numerical expression for \(F\)

A1: Correct answer (allow to the nearest penny or nearest £)