| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2016 |
| Session | June |
| Marks | 5 |
| Topic | Sequences and Series |
| Type | Fibonacci and Related Sequences |
| Difficulty | Challenging +1.8 This is a sophisticated Further Maths question requiring pattern recognition linking recursive sequences to Fibonacci numbers, formulating a conjecture from computed cases, and proving it by induction. While the individual techniques (computing terms, algebraic manipulation, induction) are standard, the conceptual leap to conjecture the Fibonacci pattern in the coefficients and the careful algebraic induction proof elevate this significantly above routine exercises. |
| Spec | 4.01a Mathematical induction: construct proofs8.01a Recurrence relations: general sequences, closed form and recurrence8.01e Fibonacci: and related sequences (e.g. Lucas numbers) |
**Question 11** [1+3+5 marks]
**(i)**
$F_3=2,\ F_4=3,\ F_5=5,\ F_6=8$ B1 (all)
**[1]**
**(ii)(a)**
$p_2(x) = 1 + \frac{1}{x+1} = \frac{x+2}{x+1}$ B1
$p_3(x) = \frac{2x+3}{x+2}$ B1
$p_4(x) = \frac{3x+5}{2x+3}$ B1 **(AG)**
**[3]**
**(b)**
$p_n(x) = \dfrac{F_n x + F_{n+1}}{F_{n-1}x + F_n}$ B1
Result is true for $n=2$ (and 3 and 4) B1 (May be mentioned later in their "round up")
Assuming $p_k(x) = \dfrac{F_k x + F_{k+1}}{F_{k-1}x + F_k}$ (not separate from their conjecture)
$p_{k+1}(x) = 1 + \dfrac{F_{k-1}x+F_k}{F_k x + F_{k+1}}$ M1
$= \dfrac{F_k x+F_{k+1}}{F_k x+F_{k+1}} + \dfrac{F_{k-1}x+F_k}{F_k x+F_{k+1}}$
$= \dfrac{(F_k+F_{k-1})x+(F_k+F_{k+1})}{F_k x+F_{k+1}}$ M1 (Collecting coefficients into successive Fib. terms)
$= \dfrac{F_{k+1}x+F_{k+2}}{F_k x+F_{k+1}}$ A1
which is the required formula with $n=k+1$.
Accept this as sufficient that proof follows by induction.
**[5]**11 (i) The sequence of Fibonacci Numbers $\left\{ F _ { n } \right\}$ is given by
$$F _ { 1 } = 1 , \quad F _ { 2 } = 1 \quad \text { and } \quad F _ { n + 1 } = F _ { n } + F _ { n - 1 } \text { for } n \geqslant 2 .$$
Write down the values of $F _ { 3 }$ to $F _ { 6 }$.\\
(ii) The sequence of functions $\left\{ \mathrm { p } _ { n } ( x ) \right\}$ is given by
$$\mathrm { p } _ { 1 } ( x ) = x + 1 \quad \text { and } \quad \mathrm { p } _ { n + 1 } ( x ) = 1 + \frac { 1 } { \mathrm { p } _ { n } ( x ) } \text { for } n \geqslant 1$$
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { p } _ { 2 } ( x )$ and $\mathrm { p } _ { 3 } ( x )$, giving each answer as a single algebraic fraction, and show that $\mathrm { p } _ { 4 } ( x ) = \frac { 3 x + 5 } { 2 x + 3 }$.
\item Conjecture an expression for $\mathrm { p } _ { n } ( x )$ as a single algebraic fraction involving Fibonacci numbers, and prove it by induction for all integers $n \geqslant 2$.
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2016 Q11 [5]}}