Year: 2010
Paper: 2
Question Number: 3
Course: LFM Pure and Mechanics
Section: Arithmetic and Geometric sequences
There were just under 1000 entries for paper II this year, almost exactly the same number as last year. Of this number, more than 60 scored over 90% while, at the other end of the scale, almost 200 failed to score more than 40 marks. In hindsight, many of the pure maths questions were a little too accessible and lacked a sufficiently tough 'difficulty gradient', so that scores were slightly higher than anticipated. This was reflected in the grade boundaries for the "1" and the "2" (around ten marks higher than is generally planned) in particular. Next year's questions may be expected to be a little bit more demanding, but only in the sense that the final 5 or 6 marks on each question should have rather more bite to them: it should certainly not be the case that all questions are tougher to get into at the outset. Most candidates attempted the requisite number of questions (six), although many of the weaker brethren made seven or eight attempts, most of which were feeble at best and they generally only picked up a maximum of 5 or 6 marks per question. It is a truth universally acknowledged that practice maketh if not perfect then at least a whole lot better prepared, and choosing to waste time on a couple of extra questions is not a good strategy on the STEPs. The major down-side of the present modular examination system is that students are not naturally prepared to approach the subject holistically; ally this to the current practice of setting highly-structured, fully-guided questions requiring no imagination, insight, depth or planning from A-level candidates in a system that fails almost nobody and rewards even the most modestly able with high grades in a manner reminiscent of a dentist giving lollipops to kids who have done little more than been brave and seen the course through, it is even more important to ensure a full and thorough preparation for these papers. The 20% of the entry who seem to be either unprepared for the rigours of a STEP, or unwittingly possessed of only a smattering of basic advanced-level skills, seems to be remarkably steady year-on-year, even in a year when their more suitably prepared compatriots found the paper appreciably easier than usual. As in previous years, the pure maths questions provided the bulk of candidates' work, with relatively few efforts to be found at the applied ones.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
The first four terms of a sequence
are given by $F_0=0$, $F_1=1$, $F_2=1$ and $F_3=2$. The general term
is given by
\[
F_n= a\lambda^n+b\mu^n\,,
\tag{$*$}
\]
where $a$, $b$, $\lambda$ and $\mu$ are independent of $n$, and $a$ is
positive.
\begin{questionparts}
\item Show that
$\lambda^2 +\lambda\mu+ \mu^2 = 2$, and find the values of
$\lambda$, $\mu$, $a$ and $b$.
\item Use $(*)$ to evaluate $F_6$.
\item Evaluate
$\displaystyle \sum_{n=0}^\infty \frac{F_n}{2^{n+1}}\,.$
\end{questionparts}\begin{questionparts}
\item $\,$
\begin{align*}
&& 0 &= a+b \tag{1}\\
&& 1 &= a\lambda -a\mu \tag{2} \\
&& 1 &= a\lambda^2 -a\mu^2 \tag{3} \\
&& 2 &= a\lambda^3 - a\mu^3 \tag{4} \\
(4) \div (2): && 2 & = \lambda^2+\lambda \mu + \mu^2 \\
(3) \div (2): && 1 &= \lambda + \mu \\
\Rightarrow && 2 &= \lambda^2 + \lambda(1-\lambda) + (1-\lambda)^2 \\
&&&= \lambda^2-\lambda+1\\
\Rightarrow && \lambda, \mu &= \frac{1 \pm \sqrt{5}}{2} \\
\Rightarrow && a &= \frac{1}{\lambda - \mu} = \frac{1}{\sqrt{5}} \\
\Rightarrow && b &= -\frac{1}{\sqrt{5}}
\end{align*}
(NB: This is Binet's formula)
\item $\,$ \begin{align*} F_6 &= \frac{1}{\sqrt{5}} \left ( \left ( \frac{1 +\sqrt{5}}{2} \right)^6- \left ( \frac{1 -\sqrt{5}}{2} \right)^6 \right) \\
&= \frac{1}{2^6 \sqrt{5}} \left ( (1+\sqrt{5})^6-(1-\sqrt{5})^6 \right) \\
&= \frac{1}{2^5 \cdot \sqrt{5}} \left (6 \sqrt{5} +\binom{6}{3} (\sqrt{5})^3+\binom{6}{5}(\sqrt{5})^5 \right)\\
&= \frac{1}{2^5} \left (6 +20\cdot 5+6\cdot 5^2 \right)\\
&= \frac{1}{2^5} 256 = 2^3 = 8
\end{align*}
(way more painful than just computing it by adding terms!)
\item $\,$
\begin{align*}
&& \sum_{n=0}^{\infty} \frac{F_n}{2^{n+1}} &= \sum_{n=0}^{\infty} \frac{a\lambda^n + b\mu^n}{2^{n+1}} \\
&&&= \frac12 \left ( \frac{a}{1-\frac{\lambda}2} + \frac{b}{1-\frac{\mu}2} \right) \\
&&&= \frac12 \left ( \frac{2a}{2-\lambda} + \frac{2b}{2-\mu}\right) \\
&&&= \frac{2a}{4-2\lambda} + \frac{2b}{4-2\mu}\\
&&&= \frac{2a}{4-(1+\sqrt{5})} - \frac{2a}{4-(1-\sqrt{5})} \\
&&&= \frac{2}{3\sqrt{5}-5} - \frac{2}{3\sqrt{5}+5} \\
&&&= \frac{6\sqrt{5}+10-6\sqrt{5}+10}{45-25} \\
&&&= 1
\end{align*}
\end{questionparts}
One doesn't need to be too devoted a mathematician to recognise the Fibonacci numbers in this question, and many candidates clearly recognised this sequence. However, they were still required to answer the question in the way specified by the wording on the paper and a lot of attempts foundered at part (ii). This was the second most frequently attempted question, yet drew the second worst marks, averaging just over 8. Most attempts got little further than (i), and many foundered even here due to a lack of appreciation of the difference of two cubes factorisation. Things clearly got much worse in (ii) when far too many folks seemed incapable of attempting a binomial expansion of (1+√5)⁶; many who did manage a decent stab at this then repeated the work for (1−√5)⁶. Very few sorted this out correctly and, as a result, there were relatively few stabs at part (iii).