| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure (Further Additional Pure) |
| Year | 2018 |
| Session | September |
| Marks | 14 |
| Topic | Sequences and series, recurrence and convergence |
| Type | Recurrence relation solving for closed form |
| Difficulty | Challenging +1.8 This Further Maths question requires solving a second-order linear recurrence relation using characteristic equations, then proving a non-trivial identity involving two sequences (essentially Pell equation solutions). The proof in part (ii)(b) requires either induction with careful algebraic manipulation or recognition of the connection to Pell's equation. The limit calculation is straightforward given the general solution. While systematic, the proof component demands sophisticated algebraic reasoning beyond standard A-level techniques. |
| Spec | 8.01d Sequence limits: limit of nth term as n tends to infinity, steady-states8.01f First-order recurrence: solve using auxiliary equation and complementary function |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Aux. Eqn. is \(\lambda^2 - 10\lambda + 1 = 0 \Rightarrow \lambda = 5 \pm \sqrt{24}\) or \(5 \pm 2\sqrt{6}\) | M1, A1 | |
| Gen. Soln. is then \(u_n = A(5 + 2\sqrt{6})^n + B(5 - 2\sqrt{6})^n\) | B1 | FT their two values of \(\lambda\) |
| (ii) (a) \(\{a_n\} = \{1, 5, \mathbf{49}, \mathbf{485}, \ldots\}\) and \(\{b_n\} = \{0, 2, \mathbf{20}, \mathbf{198}, \ldots\}\) | B1 | All highlighted terms required |
| (ii) (b) \(a_0 = 1, a_1 = 5 \Rightarrow 1 = A + B\) and \(5 = 5(A+B) + (A-B) 2\sqrt{6}\) | M1 | |
| \(\Rightarrow A = B = \frac{1}{2}\) i.e. \(a_n = \frac{1}{2}(5 + 2\sqrt{6})^n + \frac{1}{2}(5 - 2\sqrt{6})^n\) | A1 | Sim. eqns. may be solved by GC, for instance |
| \(b_0 = 0, b_1 = 2 \Rightarrow 0 = A + B\) and \(2 = 5(A+B) + (A-B) 2\sqrt{6}\) | M1 | |
| \(\Rightarrow A = -B = \frac{1}{2\sqrt{6}}\) i.e. \(b_n = \frac{1}{2\sqrt{6}}(5 + 2\sqrt{6})^n - \frac{1}{2\sqrt{6}}(5 - 2\sqrt{6})^n\) | A1 | Sim. eqns. may be solved by GC, for instance |
| Either Calling \(\alpha = 5 + 2\sqrt{6}\) and \(\beta = 5 - 2\sqrt{6}\) and substituting both in \((a_n)^2 - 6(b_n)^2 = \frac{1}{4}\{\alpha^{2n} + 2(\alpha\beta)^n + \beta^{2n}\} - 6 \times \frac{1}{24}\{\alpha^{2n} - 2(\alpha\beta)^n + \beta^{2n}\}\) | M1, A1 | The shorthand is helpful but not necessary |
| \(= (\alpha\beta)^n = 1\), by the "difference of two squares" (or product of roots) | A1 | \(\alpha\beta = 1\) must at least be stated |
| Or \((a_n)^2 - 6(b_n)^2 = (a_n - \sqrt{6}b_n)(a_n + \sqrt{6}b_n)\) | M1 | |
| \(= \frac{1}{2}\{\alpha^n + \beta^n - \alpha^n + \beta^n\} \cdot \frac{1}{2}\{\alpha^n + \beta^n + \alpha^n - \beta^n\}\) | M1 | |
| \(= (\alpha\beta)^n = 1\) as above | A1 | |
| Partial Alternative using Induction: Result true for \(n = 0, 1, 2, 3\) | B1 | Baseline case(s) ... \(\geq 2\) (consecutive) must be noted |
| Assuming \((a_{n-1})^2 - 6(b_{n-1})^2 = 1\) and \((a_n)^2 - 6(b_n)^2 = 1\) | M1 | Assuming two previous results |
| \((a_{n+1})^2 - 6(b_{n+1})^2 = (10a_n - a_{n-1})^2 - 6(10b_n - b_{n-1})^2\) | M1 | Attempt at \((n+1)\)th case via the r.r. |
| \(= 100(a_n^2 - 6b_n^2) + (a_{n-1}^2 - 6b_{n-1}^2) - 20(a_n a_{n-1} - 6b_n b_{n-1})\) | M1 | Appropriate pairing of terms |
| \(= 101 - 20(a_n a_{n-1} - 6b_n b_{n-1})\) | A1 | |
| The bracket can be addressed as above, or by an inductive proof in itself. | ||
| (ii) (c) Either For large \(n\), \((a_n)^2 \approx 6(b_n)^2 \Rightarrow \frac{a_n}{b_n} \to \sqrt{6}\) | M1, A1 | |
| Or \( | \beta | < 1 \Rightarrow a_n \to \frac{1}{2}\alpha^n\) and \(b_n \to \frac{1}{2\sqrt{6}}\alpha^n\) so that \(\frac{a_n}{b_n} \to \frac{1}{2} = \frac{\sqrt{6}}{\frac{1}{2\sqrt{6}}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| (i) If \(f^3(x) = x\) or \(f^3(a) = x\) then it will generate a group of order 2 or 3. | B1, B1 | |
| (ii) Since \(\det(\text{product}) = \text{product of dets}\) we need \(\det(M^3) = (\det M)^3 = \det(I) = 1 \Rightarrow \det(M) = \pm 1\) | E1, E1 | |
| (iii) Want \(\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} \cos\frac{2\pi}{6} & -\sin\frac{2\pi}{6} \\ \sin\frac{2\pi}{6} & \cos\frac{2\pi}{6} \end{pmatrix}\) | B1 | i.e. a generator of the group of rotations through multiples of \(\frac{\pi}{3}\) |
| (iv) (a) \(\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{3}} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}\) gives \(f(x) = \frac{\frac{1}{x} - \sqrt{3}}{\frac{\sqrt{3}}{2}x + \frac{1}{2}}\) or \(\frac{x - \sqrt{3}}{x\sqrt{3} + 1}\) | B1 | |
| (iv) (b) \(M^3 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}\) gives \(f^3(x) = \frac{-x+0}{0x-1} = x\), so \(f\) not of order 6 | B1, B1 | * BC Must state the conclusion and not stop with the working |
| (iv) (c) However, taking \(\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} \cos\frac{\pi}{6} & -\sin\frac{\pi}{6} \\ \sin\frac{\pi}{6} & \cos\frac{\pi}{6} \end{pmatrix}\) | M1, A1 | Another rotation matrix; A correct suggestion |
| gives \(f(x) = \frac{x\sqrt{3} - 1}{x + \sqrt{3}}\) for which \(f^2(x) = \frac{x - \sqrt{3}}{x\sqrt{3} + 1}\) and \(f^3(x) = \frac{-1}{x}\) \(\Rightarrow f^6(x) = x\) | M1, M1 | Attempt at powers of \(f\) up to \(f^6\) |
| Must state, or have shown, that both \(f^2(x)\) and \(f^3(x) \neq x\) | A1 |
**(i)** Aux. Eqn. is $\lambda^2 - 10\lambda + 1 = 0 \Rightarrow \lambda = 5 \pm \sqrt{24}$ or $5 \pm 2\sqrt{6}$ | M1, A1 |
Gen. Soln. is then $u_n = A(5 + 2\sqrt{6})^n + B(5 - 2\sqrt{6})^n$ | B1 | FT their two values of $\lambda$
**(ii) (a)** $\{a_n\} = \{1, 5, \mathbf{49}, \mathbf{485}, \ldots\}$ and $\{b_n\} = \{0, 2, \mathbf{20}, \mathbf{198}, \ldots\}$ | B1 | All highlighted terms required
**(ii) (b)** $a_0 = 1, a_1 = 5 \Rightarrow 1 = A + B$ and $5 = 5(A+B) + (A-B) 2\sqrt{6}$ | M1 |
$\Rightarrow A = B = \frac{1}{2}$ i.e. $a_n = \frac{1}{2}(5 + 2\sqrt{6})^n + \frac{1}{2}(5 - 2\sqrt{6})^n$ | A1 | Sim. eqns. may be solved by GC, for instance
$b_0 = 0, b_1 = 2 \Rightarrow 0 = A + B$ and $2 = 5(A+B) + (A-B) 2\sqrt{6}$ | M1 |
$\Rightarrow A = -B = \frac{1}{2\sqrt{6}}$ i.e. $b_n = \frac{1}{2\sqrt{6}}(5 + 2\sqrt{6})^n - \frac{1}{2\sqrt{6}}(5 - 2\sqrt{6})^n$ | A1 | Sim. eqns. may be solved by GC, for instance
Either **Calling** $\alpha = 5 + 2\sqrt{6}$ and $\beta = 5 - 2\sqrt{6}$ and substituting both in $(a_n)^2 - 6(b_n)^2 = \frac{1}{4}\{\alpha^{2n} + 2(\alpha\beta)^n + \beta^{2n}\} - 6 \times \frac{1}{24}\{\alpha^{2n} - 2(\alpha\beta)^n + \beta^{2n}\}$ | M1, A1 | The shorthand is helpful but not necessary
$= (\alpha\beta)^n = 1$, by the "difference of two squares" (or product of roots) | A1 | $\alpha\beta = 1$ must at least be stated
Or $(a_n)^2 - 6(b_n)^2 = (a_n - \sqrt{6}b_n)(a_n + \sqrt{6}b_n)$ | M1 |
$= \frac{1}{2}\{\alpha^n + \beta^n - \alpha^n + \beta^n\} \cdot \frac{1}{2}\{\alpha^n + \beta^n + \alpha^n - \beta^n\}$ | M1 |
$= (\alpha\beta)^n = 1$ as above | A1 |
**Partial Alternative using Induction:** Result true for $n = 0, 1, 2, 3$ | B1 | Baseline case(s) ... $\geq 2$ (consecutive) must be noted
Assuming $(a_{n-1})^2 - 6(b_{n-1})^2 = 1$ and $(a_n)^2 - 6(b_n)^2 = 1$ | M1 | Assuming two previous results
$(a_{n+1})^2 - 6(b_{n+1})^2 = (10a_n - a_{n-1})^2 - 6(10b_n - b_{n-1})^2$ | M1 | Attempt at $(n+1)$th case via the r.r.
$= 100(a_n^2 - 6b_n^2) + (a_{n-1}^2 - 6b_{n-1}^2) - 20(a_n a_{n-1} - 6b_n b_{n-1})$ | M1 | Appropriate pairing of terms
$= 101 - 20(a_n a_{n-1} - 6b_n b_{n-1})$ | A1 |
The bracket can be addressed as above, or by an inductive proof in itself. | |
**(ii) (c)** Either **For large $n$,** $(a_n)^2 \approx 6(b_n)^2 \Rightarrow \frac{a_n}{b_n} \to \sqrt{6}$ | M1, A1 |
Or $|\beta| < 1 \Rightarrow a_n \to \frac{1}{2}\alpha^n$ and $b_n \to \frac{1}{2\sqrt{6}}\alpha^n$ so that $\frac{a_n}{b_n} \to \frac{1}{2} = \frac{\sqrt{6}}{\frac{1}{2\sqrt{6}}}$ | M1, A1 |
## Question 6 Continuation (from page 8)
**(i)** If $f^3(x) = x$ or $f^3(a) = x$ then it will generate a group of order 2 or 3. | B1, B1 |
**(ii)** Since $\det(\text{product}) = \text{product of dets}$ we need $\det(M^3) = (\det M)^3 = \det(I) = 1 \Rightarrow \det(M) = \pm 1$ | E1, E1 |
**(iii)** Want $\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} \cos\frac{2\pi}{6} & -\sin\frac{2\pi}{6} \\ \sin\frac{2\pi}{6} & \cos\frac{2\pi}{6} \end{pmatrix}$ | B1 | i.e. a generator of the group of rotations through multiples of $\frac{\pi}{3}$
**(iv) (a)** $\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{3}} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}$ gives $f(x) = \frac{\frac{1}{x} - \sqrt{3}}{\frac{\sqrt{3}}{2}x + \frac{1}{2}}$ or $\frac{x - \sqrt{3}}{x\sqrt{3} + 1}$ | B1 |
**(iv) (b)** $M^3 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$ gives $f^3(x) = \frac{-x+0}{0x-1} = x$, so $f$ not of order 6 | B1, B1 | * BC Must state the conclusion and not stop with the working
**(iv) (c)** However, taking $\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} \cos\frac{\pi}{6} & -\sin\frac{\pi}{6} \\ \sin\frac{\pi}{6} & \cos\frac{\pi}{6} \end{pmatrix}$ | M1, A1 | Another rotation matrix; A correct suggestion
gives $f(x) = \frac{x\sqrt{3} - 1}{x + \sqrt{3}}$ for which $f^2(x) = \frac{x - \sqrt{3}}{x\sqrt{3} + 1}$ and $f^3(x) = \frac{-1}{x}$ $\Rightarrow f^6(x) = x$ | M1, M1 | Attempt at powers of $f$ up to $f^6$
Must state, or have shown, that both $f^2(x)$ and $f^3(x) \neq x$ | A1 |The members of the family of the sequences $\{u_n\}$ satisfy the recurrence relation
$$u_{n+1} = 10u_n - u_{n-1} \text{ for } n \geq 1. \quad (*)$$
\begin{enumerate}[label=(\roman*)]
\item Determine the general solution of (*). [3]
\item The sequences $\{a_n\}$ and $\{b_n\}$ are members of this family of sequences, corresponding to the initial terms $a_0 = 1$, $a_1 = 5$ and $b_0 = 0$, $b_1 = 2$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Find the next two terms of each sequence. [1]
\item Prove that, for all non-negative integers $n$, $(a_n)^2 - 6(b_n)^2 = 1$. [8]
\item Determine $\lim_{n \to \infty} \frac{a_n}{b_n}$. [2]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure 2018 Q7 [14]}}