Year: 2020
Paper: 2
Question Number: 3
Course: UFM Additional Further Pure
Section: Sequences and Series
No solution available for this problem.
There were just over 800 entries for this paper, and good solutions were seen to all of the questions. Candidates should be aware of the need to provide clear explanations of their reasoning throughout the paper (and particularly in questions where the result to be shown is given in the question). Short explanatory comments at key points in solutions can be very helpful in this regard, as can clearly drawn diagrams of the situation described in the question. The paper included a few questions where a statement of the form "A if and only if B" needed to be proven – candidates should be aware of the meaning of such statements and make sure that both directions of the implication are covered clearly. In general, candidates who performed better on the questions in this paper recognised the relationships between the different parts of each question and were able to adapt methods used in earlier parts when working on the later sections of the question.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
A sequence $u_1, u_2, \ldots, u_n$ of positive real numbers is said to be unimodal if there is a value $k$ such that
\[u_1 \leqslant u_2 \leqslant \ldots \leqslant u_k\]
and
\[u_k \geqslant u_{k+1} \geqslant \ldots \geqslant u_n.\]
So the sequences $1, 2, 3, 2, 1$;\ $1, 2, 3, 4, 5$;\ $1, 1, 3, 3, 2$ and $2, 2, 2, 2, 2$ are all unimodal, but $1, 2, 1, 3, 1$ is not.
A sequence $u_1, u_2, \ldots, u_n$ of positive real numbers is said to have property $L$ if $u_{r-1}u_{r+1} \leqslant u_r^2$ for all $r$ with $2 \leqslant r \leqslant n-1$.
\begin{questionparts}
\item Show that, in any sequence of positive real numbers with property $L$,
\[u_{r-1} \geqslant u_r \implies u_r \geqslant u_{r+1}.\]
Prove that any sequence of positive real numbers with property $L$ is unimodal.
\item A sequence $u_1, u_2, \ldots, u_n$ of real numbers satisfies $u_r = 2\alpha u_{r-1} - \alpha^2 u_{r-2}$ for $3 \leqslant r \leqslant n$, where $\alpha$ is a positive real constant. Prove that, for $2 \leqslant r \leqslant n$,
\[u_r - \alpha u_{r-1} = \alpha^{r-2}(u_2 - \alpha u_1)\]
and, for $2 \leqslant r \leqslant n-1$,
\[u_r^2 - u_{r-1}u_{r+1} = (u_r - \alpha u_{r-1})^2.\]
Hence show that the sequence consists of positive terms and is unimodal, provided $u_2 > \alpha u_1 > 0$.
In the case $u_1 = 1$ and $u_2 = 2$, prove by induction that $u_r = (2-r)\alpha^{r-1} + 2(r-1)\alpha^{r-2}$.
Let $\alpha = 1 - \dfrac{1}{N}$, where $N$ is an integer with $2 \leqslant N \leqslant n$.
In the case $u_1 = 1$ and $u_2 = 2$, prove that $u_r$ is largest when $r = N$.
\end{questionparts}
This was the second least attempted of the pure questions. Relatively few candidates made a complete attempt at all of the parts and only 4 achieved full marks for the question. The question consisted of a succession of given results which were to be established. Thus, candidates needed to be more aware of the importance of providing careful and thorough explanations and justifications for each step that they took along the way. Many marks were lost as a result of carelessness in providing all of the necessary details. A significant number of candidates thought that the implication in (i) showed that the sequence was either increasing or decreasing and so got little or no credit. Establishing the given relations in (i) was generally done quite well, with candidates demonstrating a considerable range of algebraic skills in their working. But then a lot of candidates failed to show that the sequence was positive, which undermined their attempts to deduce that the sequence was unimodal. Many candidates used an induction proof for the first proof in part (ii) despite the fact that a more direct approach was possible and considerably simpler. The "asked-for" induction proof was usually handled well, though establishing the baseline case was often flawed; many either overlooked the need to establish both of the cases n = 1 and n = 2 or, when giving a one-step induction proof with the help of the previously-established result, chose an incorrect baseline case. The final part of the question was often avoided, though full attempts often gained full credit. Again, the usual oversight was to fail to establish positivity. Many of those who produced only a faltering solution here overlooked the need to compare successive terms, usually merely working with an expression for ur only, often with the use of differentiation - attempts along such lines invariably lost all of the final 7 marks allocated, primarily because the required result is based on discrete values of r while calculus works with continuous values.