| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure (Further Additional Pure) |
| Year | 2018 |
| Session | December |
| Marks | 13 |
| Topic | Sequences and Series |
| Type | First-Order Linear Recurrence Relations |
| Difficulty | Challenging +1.2 This is a Further Maths question on first-order linear recurrence relations. Part (a) is routine calculation, part (b) is standard technique for solving recurrence relations, but part (c) requires proof by induction and number theory reasoning about modular arithmetic to show no term is a perfect square. The multi-step nature and proof requirements elevate it above average difficulty, but the techniques are well-established for Further Maths students. |
| Spec | 1.04e Sequences: nth term and recurrence relations4.01a Mathematical induction: construct proofs8.01a Recurrence relations: general sequences, closed form and recurrence8.01f First-order recurrence: solve using auxiliary equation and complementary function |
| Answer | Marks | Guidance |
|---|---|---|
| \(u_2 = 13, u_3 = 53, u_4 = 213\) | B1 | 1.1 |
| Total: [1] | All three |
| Answer | Marks | Guidance |
|---|---|---|
| Auxiliary Equation is \(m - 4 = 0\) | B1 | 1.2 |
| so that Complementary Solution (CS) is \(u_n = A(4^n)\) | B1 | 1.1 |
| Particular Solution (PS) is \(u_o = a\) with \(a - 4a = 1\) i.e. PS is \(u_n = -\frac{1}{3}\) | B1 | 1.1 |
| General Solution is \(u_n = A(4^n) - \frac{1}{3}\) | B1FT | 1.1 |
| Using \(n = 1, u_1 = 3\), to find the value of \(A\) | M1 | 1.1a |
| \(u_n = \frac{5}{6}(4^n) - \frac{1}{3}\) or equivalent | A1 | 1.1 |
| Total: [5] | FT provided CS has 1 arbitrary constant, PS none soi; or \(A = \frac{5}{6}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Result is true for \(n = 1\) since \(u_1 = 10 \times 0 + 3 = 3\) | B1 | 2.2a |
| Assuming \(u_k = 10m + 3\) | M1 | 2.1 |
| Then \(u_{k+1} = 4(10m + 3) + 1 =\) | M1 | 1.1a |
| \(10(4m + 1) + 3\) | A1 | 2.5 |
| \(\Rightarrow\) result true for \(n = k + 1\) when true for \(n = k\); since true for \(n = 1\), it follows that it is true for all positive integers \(n\) by induction | E1 | 2.4 |
| Total: [5] | Suitable 'round-up' of the induction logic |
| Answer | Marks | Guidance |
|---|---|---|
| E.g. The quadratic residues mod 10 are 0, 1, 4, 9, 5, 6 | M1 | 3.1a |
| All \(u_n\)'s are ≡ 3 (mod 10) and hence not squares | A1 | 3.2a |
| Total: [2] | Or listing units digits of squares in base 10. May note quadratic non-residues 2, 3, 7 and 8 |
**Part (a)(i)**
| $u_2 = 13, u_3 = 53, u_4 = 213$ | B1 | 1.1 |
| Total: [1] | All three | |
**Part (a)(ii)**
| Auxiliary Equation is $m - 4 = 0$ | B1 | 1.2 |
| so that Complementary Solution (CS) is $u_n = A(4^n)$ | B1 | 1.1 |
| Particular Solution (PS) is $u_o = a$ with $a - 4a = 1$ i.e. PS is $u_n = -\frac{1}{3}$ | B1 | 1.1 |
| General Solution is $u_n = A(4^n) - \frac{1}{3}$ | B1FT | 1.1 |
| Using $n = 1, u_1 = 3$, to find the value of $A$ | M1 | 1.1a |
| $u_n = \frac{5}{6}(4^n) - \frac{1}{3}$ or equivalent | A1 | 1.1 |
| Total: [5] | FT provided CS has 1 arbitrary constant, PS none soi; or $A = \frac{5}{6}$ | |
**Part (b)(i)**
| Result is true for $n = 1$ since $u_1 = 10 \times 0 + 3 = 3$ | B1 | 2.2a |
| Assuming $u_k = 10m + 3$ | M1 | 2.1 |
| Then $u_{k+1} = 4(10m + 3) + 1 =$ | M1 | 1.1a |
| $10(4m + 1) + 3$ | A1 | 2.5 |
| $\Rightarrow$ result true for $n = k + 1$ when true for $n = k$; since true for $n = 1$, it follows that it is true for all positive integers $n$ by induction | E1 | 2.4 |
| Total: [5] | Suitable 'round-up' of the induction logic | |
**Part (b)(ii)**
| E.g. The quadratic residues mod 10 are 0, 1, 4, 9, 5, 6 | M1 | 3.1a |
| All $u_n$'s are ≡ 3 (mod 10) and hence not squares | A1 | 3.2a |
| Total: [2] | Or listing units digits of squares in base 10. May note quadratic non-residues 2, 3, 7 and 8 | |2 A sequence $\left\{ u _ { n } \right\}$ is given by $u _ { n + 1 } = 4 u _ { n } + 1$ for $n \geqslant 1$ and $u _ { 1 } = 3$.
\begin{enumerate}[label=(\alph*)]
\item Find the values of $u _ { 2 } , u _ { 3 }$ and $u _ { 4 }$.
\item Solve the recurrence system (*).
\item \begin{enumerate}[label=(\roman*)]
\item Prove by induction that each term of the sequence can be written in the form $( 10 m + 3 )$ where $m$ is an integer.
\item Show that no term of the sequence is a square number.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure 2018 Q2 [13]}}