Year: 2023
Paper: 3
Question Number: 4
Course: LFM Stats And Pure
Section: Binomial Theorem (positive integer n)
No solution available for this problem.
The total entry was a marginal increase on that of 2022 (by just over 1%). Two questions were attempted by more than 90% of candidates, another two by 80%, and another two by about two thirds. The least popular questions were attempted by more than a sixth of candidates. All the questions were perfectly answered by at least three candidates (but mostly more than this), with one being perfectly answered by eighty candidates. Very nearly 90% of candidates attempted no more than 7 questions. One general comment regarding all the questions is that candidates need to make sure that they read the question carefully, paying particular attention to command words such as "hence" and "show that".
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
Let $n$ be a positive integer. The polynomial $\mathrm{p}$ is defined by the identity
\[\mathrm{p}(\cos\theta) \equiv \cos\big((2n+1)\theta\big) + 1\,.\]
\begin{questionparts}
\item Show that
\[\cos\big((2n+1)\theta\big) = \sum_{r=0}^{n} \binom{2n+1}{2r} \cos^{2n+1-2r}\theta\,(\cos^2\theta - 1)^r\,.\]
\item By considering the expansion of $(1+t)^{2n+1}$ for suitable values of $t$, show that the coefficient of $x^{2n+1}$ in the polynomial $\mathrm{p}(x)$ is $2^{2n}$.
\item Show that the coefficient of $x^{2n-1}$ in the polynomial $\mathrm{p}(x)$ is $-(2n+1)2^{2n-2}$.
\item It is given that there exists a polynomial $\mathrm{q}$ such that
\[\mathrm{p}(x) = (x+1)\,[\mathrm{q}(x)]^2\]
and the coefficient of $x^n$ in $\mathrm{q}(x)$ is greater than $0$.
Write down the coefficient of $x^n$ in the polynomial $\mathrm{q}(x)$ and, for $n \geqslant 2$, show that the coefficient of $x^{n-2}$ in the polynomial $\mathrm{q}(x)$ is
\[2^{n-2}(1-n)\,.\]
\end{questionparts}
This was only marginally more popular than question 3 and was the least successfully attempted question on the paper with a mean score of 5/20. In the vast majority of cases, there was no substantially correct attempt except in part (i). Those that used de Moivre's theorem, expanded binomially, equated real parts and replaced sin²ʳθ by (cos²θ − 1)ʳ, generally scored well in this part, but marks could not be credited where mathematical steps were glossed over when the question stated, 'Show that'. Some attempted use of proof by induction, but their conclusions were not supported by their mathematical argument. Many attempting part (ii) wrote down the coefficient required but made no further progress, or made the expansion and went no further. There were some that did substitute +1 and -1, and solved this part. Part (iii) was solved generally by those that had made the substitutions in (ii) and saw that differentiation might be useful. Part (iv) could be answered using the given results from the previous parts of the question, however this part was almost exclusively only attempted by candidates that had had a reasonable level of success on the previous three parts. Those candidates that set out a careful and organised solution were more successful in part (iv).