Year: 2012
Paper: 2
Question Number: 1
Course: LFM Stats And Pure
Section: Generalised Binomial Theorem
There were just over 1000 entries for paper II this year, almost exactly the same number as last year. Overall, the paper was found marginally easier than its predecessor, which means that it was pitched at exactly the level intended and produced the hoped-for outcomes. Almost 50 candidates scored 100 marks or more, with more than 400 gaining at least half marks on the paper. At the lower end of the scale, around a quarter of the entry failed to score more than 40 marks. It was pleasing to note that the advice of recent years, encouraging students not to make attempts at lots of early parts to questions but rather to spend their time getting to grips with the six that can count towards their paper total, was more obviously being heeded in 2012 than I can recall being the case previously. 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. Questions 1 and 2 were the most popular questions, although each drew only around 800 "hits" – fewer than usual. Questions 3 – 5 & 8 were almost as popular (around 700), with Q6 attracting the interest of under 450 candidates and Q7 under 200. Q9 was the most popular applied question – and, as it turned out, the most successfully attempted question on the paper – with very little interest shown in the rest of Sections B or C.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
Write down the general term in the expansion in powers of $x$ of $(1-x^6)^{-2}\,$.
\begin{questionparts}
\item
Find the coefficient of $x^{24}$ in
the expansion in powers of $x$ of
\[
(1-x^6)^{-2} (1-x^3)^{-1}\,.\]
Obtain also, and simplify, formulae for the
coefficient of $x^n$ in the different
cases that arise.
\item Show that the coefficient of $x^{24}$
in the expansion in powers of $x$ of
\[
(1-x^6)^{-2} (1-x^3)^{-1} (1-x)^{-1}\,\] is $55$,
and find the coefficients of
$x^{25}$ and $x^{66}$.
\end{questionparts}$\displaystyle (1-x^6)^{-2} = \sum_{n=0}^{\infty} (n+1)x^{6n}$
\begin{questionparts}
\item $\,$ \begin{align*}
&& f(x) &= (1-x^6)^{-2}(1-x^3)^{-1} \\
&&&= \left ( \sum_{n=0}^{\infty} (n+1)x^{6n} \right) \left ( \sum_{n=0}^{\infty} x^{3n} \right) \\
[x^{24}]: && c_{24} &= 1 + 2+ 3+4+5 = 15
\end{align*}
Clearly $n$ must be a multiple of $3$.
If $n = 6k$ then we have $1 + 2 + \cdots + (k+1) = \frac{(k+1)(k+2)}{2}$
If $n = 6k+3$ then we have $1 + 2 + \cdots + (k+1) = \frac{(k+1)(k+2)}{2}$ the same way, we just must always get one extra $x^3$ term from the second expansion.
\item We can obtain $x^{24}$ from the product of $(1-x^6)^{-2}(1-x^3)^{-1}$ and $(1-x)^{-1}$ in the following ways:
\begin{array}{c|c|c}
(1-x^6)^{-2}(1-x^3)^{-1} & (1-x)^{-1} & \text{product} \\ \hline
15x^{24} & x^0 & 15x^{24} \\
10x^{21} & x^3 & 10x^{24} \\
10x^{18} & x^6 & 10x^{24} \\
6x^{15} & x^9 & 6x^{24} \\
6x^{12} & x^{12} & 6x^{24} \\
3x^{9} & x^{15} & 3x^{24} \\
3x^{6} & x^{18} & 3x^{24} \\
x^{3} & x^{21} & x^{24} \\
x^{0} & x^{24}& x^{24}
\end{array}
So the total is $55$.
Similarly for $25$ we can only obtain this in the same ways but also taking an extra power of $x$ from the geometric series, ie $55$
For $66$ we obtain by similar reasoning that it is:
$\frac{13\cdot12}{2} + 2 \left (1 + 3 + \cdots + \frac{13 \cdot 12}{2} \right) = \frac{13\cdot12}{2} + 2 \binom{14}{3} = \frac{13 \cdot 12}2 ( 1 + \frac{30}{3}) = 11 \cdot 6 \cdot 13 = 858$
\end{questionparts}
The first question is set with the intention that everyone should be able to attempt it, but 20% of candidates were clearly put off by the algebraic nature of this year's opener. It was, nonetheless, the most popular question on the paper, possibly due to the lack of any advanced techniques. The specifically numerical parts of the question were generally more confidently, and hence successfully, handled than the general ones. So, for instance, formulae for the various coefficients in (i) were often not correct, even when high marks were scored on the question. When numerical answers went astray, it was usually due to incorrect signs in the early stages, with candidates failing to realise that all terms were positive. The very final demand (for the coefficient of x66) was the real test, not only of candidates' resilience and nerve but also of their grasp of where the various contributions were coming from. From a marking point of view, very few candidates gave particularly clear methods, and it was usually difficult for the markers to decipher the underlying processes from what appeared to be merely a whole load of numbers written down and added up.