Year: 2022
Paper: 2
Question Number: 3
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
The Fibonacci numbers are defined by $F_0 = 0$, $F_1 = 1$ and, for $n \geqslant 0$, $F_{n+2} = F_{n+1} + F_n$.
\begin{questionparts}
\item Prove that $F_r \leqslant 2^{r-n} F_n$ for all $n \geqslant 1$ and all $r \geqslant n$.
\item Let $S_n = \displaystyle\sum_{r=1}^{n} \frac{F_r}{10^r}$.
Show that
\[\sum_{r=1}^{n} \frac{F_{r+1}}{10^{r-1}} - \sum_{r=1}^{n} \frac{F_r}{10^{r-1}} - \sum_{r=1}^{n} \frac{F_{r-1}}{10^{r-1}} = 89S_n - 10F_1 - F_0 + \frac{F_n}{10^n} + \frac{F_{n+1}}{10^{n-1}}\,.\]
\item Show that $\displaystyle\sum_{r=1}^{\infty} \frac{F_r}{10^r} = \frac{10}{89}$ and that $\displaystyle\sum_{r=7}^{\infty} \frac{F_r}{10^r} < 2 \times 10^{-6}$. Hence find, with justification, the first six digits after the decimal point in the decimal expansion of $\dfrac{1}{89}$.
\item Find, with justification, a number of the form $\dfrac{r}{s}$ with $r$ and $s$ both positive integers less than $10000$ whose decimal expansion starts
\[0.0001010203050813213455\ldots\,.\]
\end{questionparts}
There were several good attempts to this question, particularly in the middle section. However, the induction required in the first part of the question caused difficulties for many candidates. In particular, many did not realise which variable it was necessary to perform the induction on. Attempts to the second part of the question were much better and candidates were generally able to demonstrate a good level of skill in manipulating summations, including changing the variable over which the sum ranges. Unfortunately, some slips in the algebra were seen in several cases, but many candidates were able to reach the given result convincingly. Many candidates recognised that the left-hand side of the previous result must be equal to zero and were therefore able to show the values of the sums successfully. Some care was needed with the explanation of how the second sum led to certainty about the first six digits of the expansion. Candidates who had successfully completed part (iii) of the question were often then able to identify the correct approach for the final part. However, the justifications often failed to calculate the value of the error when only considering some of the terms of the sequence.