A2 June 2025 Q7
7. A sequence \(\{u_n\}\), where \(n \geqslant 0\), satisfies the first order recurrence relation (*)
\[u_{n+1} = \alpha u_n - \beta\left(k^n\right) \qquad (*)\]where \(\alpha\), \(\beta\), \(k\) are non-zero constants and \(\alpha \neq k\)
Given that \(u_0 = L\) and \(u_T = 0\), where \(T\) is a constant,
Taylor borrows a sum of money, \(L\), to buy a house.
Taylor arranges a loan which enables them to increase their repayment as their income grows.
The terms of the loan require Taylor’s repayments to be made once a year.
On each anniversary of the loan being taken out, 6% interest is charged on the outstanding balance and Taylor makes an annual repayment of the total loan.
Taylor’s total annual repayment of the loan will increase by 4% each year and the whole debt will be cleared in exactly 40 years.
The recurrence relation (*) can be used to model this situation.
| Scheme | Marks | AO |
|---|---|---|
| aux. equation is \(m - \alpha = 0 \Rightarrow\) complementary function is \(A\alpha^n\) | B1 | 2.1 |
| Particular solution try \(u_n = Bk^n\) and substitute into recurrence relation | M1 | 1.1b |
| \(Bk^{n+1} - \alpha Bk^n = -\beta k^n \Rightarrow Bk - \alpha B = -\beta\) and solve for \(B\) | dM1 | 1.1b |
| \(u_n = A\alpha^n + \dfrac{\beta k^n}{\alpha - k}\) | A1 | 2.2a |
| (4) |
Notes
B1: CAO for complementary function (condone if stated as \(u_n = A\alpha^n\))
M1: correct form for particular solution and substituted into recurrence relation
dM1: simplifying (by eliminating \(n\)) and solving for their constant (\(B\))
A1: correct general solution (must be \(u_n =\) ) (must be fully simplified)
| Scheme | Marks | AO |
|---|---|---|
| From given conditions: \(u_0 = L \Rightarrow A + \dfrac{\beta}{\alpha - k} = L\) and \(u_T = 0 \Rightarrow A\alpha^T + \dfrac{\beta k^T}{\alpha - k} = 0\) | M1 | 3.4 |
| Eliminating \(\beta\) which if correct is \(A\alpha^T + (L - A)k^T = 0\) | dM1 | 3.1a |
| Re-arrange to get an expression for \(A\) in terms of \(L\), \(k\), \(T\) and \(\alpha\) (which for reference if correct is \(A = \dfrac{Lk^T}{k^T - \alpha^T}\)) | ddM1 | 3.1a |
| Eliminate \(A\) and \(\beta\) from their general solution to obtain an expression for \(u_n\) in terms of \(L\), \(k\), \(T\), \(n\) and \(\alpha\) only \(u_n = \left(\dfrac{Lk^T}{k^T - \alpha^T}\right)\alpha^n + \left(L - \dfrac{Lk^T}{k^T - \alpha^T}\right)k^n\) | dddM1 | 1.1b |
| \(u_n = \dfrac{Lk^n}{\alpha^T - k^T}\left(\alpha^T - k^{T-n}\alpha^n\right) = \dfrac{Lk^n}{\frac{\alpha^T}{\alpha^T} - \frac{k^T}{\alpha^T}}\left(\dfrac{\alpha^T}{\alpha^T} - \dfrac{k^{T-n}\alpha^n}{\alpha^T}\right)\) \(u_n = \dfrac{Lk^n}{1 - \left(\frac{k}{\alpha}\right)^T}\left(1 - \left(\frac{k}{\alpha}\right)^{T-n}\right)\) * | A1 | 2.2a |
| (5) |
Notes
M1: Form two equations using their general solution, and both given conditions
dM1: Eliminate either \(A\) or \(\beta\) from their two equations (formed from the given conditions)
ddM1: Re-arrange to obtain an expression for either \(\beta\) or \(A\) in terms of \(L\), \(k\), \(T\), \(n\) and \(\alpha\)
They may eliminate \(A\) and then obtain an expression for \(\dfrac{\beta}{\alpha - k}\) and then substitute for this
dddM1: Eliminate both \(A\) and \(\beta\) from their general solution to obtain an expression for \(u_n\) in terms of \(L\), \(k\), \(T\), \(n\) and \(\alpha\) only
A1: CSO – correctly derive given result (so sufficient working must be shown)
In part (b) for those that at attempt to verify (rather than show the given result) then the first 3 M marks are as in the main scheme (M1 for using the given conditions to form two equations, dM1 for eliminating \(A\), ddM1 for obtaining an expression for \(\beta\) (which for reference if correct is \(\beta = \dfrac{L\alpha^T(\alpha - k)}{\alpha^T - k^T}\)), then the fourth M mark is for obtaining an expression for \(u_{n+1} - \alpha u_n\) in terms of \(L\), \(k\), \(T\), \(n\) and \(\alpha\) only – which if correct should be equivalent to \(u_{n+1} - \alpha u_n = \dfrac{Lk^n}{1 - \left(\frac{k}{\alpha}\right)^T}\left\{k - k\left(\frac{k}{\alpha}\right)^{T-n}\left(\dfrac{\alpha}{k}\right) - \alpha + \alpha\left(\frac{k}{\alpha}\right)^{T-n}\right\} = -k^n\left\{\dfrac{L(\alpha - k)}{1 - \left(\frac{k}{\alpha}\right)^T}\right\}\). The final A mark is for correctly showing that \(\dfrac{L(\alpha - k)}{1 - \left(\frac{k}{\alpha}\right)^T}\) is equivalent to \(\dfrac{L\alpha^T(\alpha - k)}{\alpha^T - k^T}\) (so explicitly showing that \(u_{n+1} - \alpha u_n\) is equal to \(-\beta(k^n)\) with a correct expression for \(\beta\) explicitly seen (in terms of \(L\), \(k\), \(T\) and \(\alpha\))). (corrected from the printed mark scheme: the printed note says “In part (c)”; it refers to part (b))
Alternative for (b)
If they eliminate \(A\) first
\(\beta = \dfrac{L\alpha^T(\alpha - k)}{(\alpha^T - k^T)}\) and \(A = L - \dfrac{L\alpha^T}{(\alpha^T - k^T)} = \dfrac{-Lk^T}{\alpha^T - k^T}\)
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\alpha = 1.06\) (ii) \(k = 1.04\) (iii) \(\beta\) is the amount that Taylor repaid on the first anniversary of the loan being taken out | B1 B1 B1 | 2.2a 2.2a 3.1b |
| (3) |
Notes
(i) B1: CAO for \(\alpha\)
(ii) B1: CAO for \(k\)
(iii) B1: CAO for the relevance of \(\beta\) in (*) must be clear that this is the amount repaid one year after the loan is taken out. Accept e.g original amount repaid or initial repayment and condone reference to 0th year if clear that this is the first repayment (do not accept annual repayment)
| Scheme | Marks | AO |
|---|---|---|
| \(u_{30} = \dfrac{L(1.04)^{30}}{1 - \left(\frac{1.04}{1.06}\right)^{40}}\left[1 - \left(\dfrac{1.04}{1.06}\right)^{40-30}\right]\) | M1 | 3.4 |
| \(u_{30} = (1.05\ldots)L > L\) so, after 30 years the outstanding debt (which is \((1.05\ldots)L\)) is larger than the original loan (\(L\)) | A1 | 3.2a |
| (2) | ||
| (14 marks) |
Notes
M1: Substitute \(T = 40\), \(n = 30\) and their values of \(\alpha, k\) into correct particular solution (accept sight of \(1.05L\) for this mark even if full calculation not shown)
A1: CAO that the outstanding debt after 30 years is larger than the original debt (so must see some reference to \(1.05L\) (or better) and \(L\)) (accept \(1.05L > L\)) (must follow from a correct calculation)