Year: 2022
Paper: 2
Question Number: 2
Course: UFM Additional Further Pure
Section: Sequences and Series
No solution available for this problem.
Candidates appeared to be generally well prepared for most topics within the examination, but there were a few situations in questions where some did not appear to be as proficient in standard techniques as needed. In particular, the method for finding invariant lines required in question 8 and the manipulation of trigonometric functions that were needed in question 10 caused considerable difficulties for some candidates. An additional issue that occurred at numerous points in the paper relates to the direction in which a deduction is required. It is important that candidates make sure that they know which statement is the one that they should start from as they deduce the other and that it is clear in their solution that the logic has gone in the correct direction. Clarity of solution is also an issue that candidates should be aware of, especially in the situations where the result to be reached has been given. It is important to check that there are no special cases that need to be considered separately, and when dividing by a function it is necessary to confirm that the function cannot be equal to 0 (and in the case of inequalities that the function always has the same sign). When drawing diagrams and sketching graphs it is useful if significant points that need to be clear are not drawn over the lines on the page as these can be difficult to interpret during the marking process.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
A sequence $u_n$, where $n = 1, 2, \ldots$, is said to have \emph{degree} $d$ if $u_n$, as a function of $n$, is a polynomial of degree $d$.
\begin{questionparts}
\item Show that, in any sequence $u_n$ $(n = 1, 2, \ldots)$ that satisfies $u_{n+1} = \frac{1}{2}(u_{n+2} + u_n)$ for all $n \geqslant 1$, there is a constant difference between successive terms.
Deduce that any sequence $u_n$ for which $u_{n+1} = \frac{1}{2}(u_{n+2} + u_n)$, for all $n \geqslant 1$, has degree at most 1.
\item The sequence $v_n$ $(n = 1, 2, \ldots)$ satisfies $v_{n+1} = \frac{1}{2}(v_{n+2} + v_n) - p$ for all $n \geqslant 1$, where $p$ is a non-zero constant. By writing $v_n = t_n + pn^2$, show that the sequence $v_n$ has degree 2.
Given that $v_1 = v_2 = 0$, find $v_n$ in terms of $n$ and $p$.
\item The sequence $w_n$ $(n = 1, 2, \ldots)$ satisfies $w_{n+1} = \frac{1}{2}(w_{n+2} + w_n) - an - b$ for all $n \geqslant 1$, where $a$ and $b$ are constants with $a \neq 0$. Show that the sequence $w_n$ has degree 3.
Given that $w_1 = w_2 = 0$, find $w_n$ in terms of $n$, $a$ and $b$.
\end{questionparts}
Many candidates who attempted this question struggled to achieve high marks as there were several points within this question where the reasoning required careful explanation. Part (i) was often completed successfully by candidates who recognised that the relationship given could easily be rearranged to show the required result. Approaches which did not recognise this and tried to use the relationship expressed for different triples of terms were unable to make any significant progress. Candidates who attempted to argue that higher powers would not lead to a common difference were generally not successful in showing that the sequence has degree at most one. The first section of part (ii) was generally well done by those who attempted it, and the relationship with the first part was often seen although in some cases candidates omitted to observe that p was not equal to zero. Candidates were also often able to deduce the formula for the sequence, either by substituting a general form or by looking at the differences between terms. Many of the candidates who attempted part (iii) recognised that a similar approach to part (ii) would be likely to work. However, many assumed that the required coefficient of n³ would be a, rather than using a variable so that the correct coefficient could be deduced at a later point. The algebra for this part was more complicated and some struggled to follow through the work accurately. Having reached the point where the correct value of k could be deduced it was then necessary to consider when the new sequence would be of the form in part (ii) and when it would be of the form in part (i) and the analysis of these cases was not always completed fully.