Year: 2023
Paper: 2
Question Number: 5
Course: UFM Pure
Section: Sequences and series, recurrence and convergence
No solution available for this problem.
Many candidates were able to express their reasoning clearly and presented good solutions to the questions that they attempted. There were excellent solutions seen for all of the questions. An area where candidates struggled in several questions was in the direction of the logic that was required in a solution. Some candidates failed to appreciate that separate arguments may be needed for the "if" and "only if" parts of a question and, in some cases, candidates produced correct arguments, but for the wrong direction. In several questions it was clear that candidates who used sketches or diagrams generally performed much better that those who did not. Sketches often also helped to make the solution clearer and easier to understand. Several questions on the STEP papers ask candidates to show a given result. Candidates should be aware that there is a need to present sufficient detail in their solutions so that it is clear that the reasoning is well understood.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\begin{questionparts}
\item The sequence $x_n$ for $n = 0, 1, 2, \ldots$ is defined by $x_0 = 1$ and by
\[x_{n+1} = \frac{x_n + 2}{x_n + 1}\]
for $n \geqslant 0$.
\begin{enumerate}
\item[(a)] Explain briefly why $x_n \geqslant 1$ for all $n$.
\item[(b)] Show that $x_{n+1}^2 - 2$ and $x_n^2 - 2$ have opposite sign, and that
\[\left|x_{n+1}^2 - 2\right| \leqslant \tfrac{1}{4}\left|x_n^2 - 2\right|\,.\]
\item[(c)] Show that
\[2 - 10^{-6} \leqslant x_{10}^2 \leqslant 2\,.\]
\end{enumerate}
\item The sequence $y_n$ for $n = 0, 1, 2, \ldots$ is defined by $y_0 = 1$ and by
\[y_{n+1} = \frac{y_n^2 + 2}{2y_n}\]
for $n \geqslant 0$.
\begin{enumerate}
\item[(a)] Show that, for $n \geqslant 0$,
\[y_{n+1} - \sqrt{2} = \frac{(y_n - \sqrt{2})^2}{2y_n}\]
and deduce that $y_n \geqslant 1$ for $n \geqslant 0$.
\item[(b)] Show that
\[y_n - \sqrt{2} \leqslant 2\left(\frac{\sqrt{2}-1}{2}\right)^{2^n}\]
for $n \geqslant 1$.
\item[(c)] Using the fact that
\[\sqrt{2} - 1 < \tfrac{1}{2}\,,\]
or otherwise, show that
\[\sqrt{2} \leqslant y_{10} \leqslant \sqrt{2} + 10^{-600}\,.\]
\end{enumerate}
\end{questionparts}
In part (i)(a) most candidates realised that induction was necessary. Although "explain briefly" was written in the question, some candidates omitted necessary components of an inductive argument here. Some candidates incorrectly stated that the sequence always increased. A popular alternative method was stating xn+2 > xn+1. In this case it is necessary to observe that the denominator is positive to secure full marks. In part (i)(b) many candidates were successful here in rewriting xn+1²-2 in terms of xn but some failed to assert (and very briefly justify) the strict positivity of (xn+1)² in order to show that xn+1²-2 and xn²-2 have opposite signs. When showing |xn+1²-2| < |xn²-2|/4 the most common mistake was to not use absolute value signs, and write false assertions like xn+1²-2 < (xn²-2)/4, which is false for odd n. In part (i)(c) many students used the inequality in the previous part repeatedly to write |x10²-2| < |x0²-2|/4^10 but did not give a justification that 4^10 > 10^6. A small number of candidates were able to calculate x10, and x10² successfully and numerically compare these to 2 and 2-10^-6, however almost all attempts at this were unsuccessful. Almost all candidates who attempted part (ii)(a) earned at least one mark. In several cases candidates did not formulate a standard inductive argument, either missing the base case or not using an inductive hypothesis. In part (ii)(b) many candidates used n=0 as a base case, but this is not valid here. Of those who opted for an alternative method of using recursion to write yn - √2 in terms of y0 - √2, few were able to justify the exponent for powers of 2. Candidates who attempted a full inductive proof often earned at least 2 of the 4 marks for this part. Candidates attempting part (ii)(c) often earned some marks for showing the correct method, but errors in the accuracy of the work meant that few were able to achieve full marks here.