Year: 2017
Paper: 1
Question Number: 8
Course: LFM Pure
Section: Proof by induction
The pure questions were again the most popular of the paper with questions 1 and 3 being attempted by almost all candidates. The least popular questions on the paper were questions 10, 11, 12 and 13 and a significant proportion of attempts at these were brief, attracting few or no marks. Candidates generally demonstrated a high level of competence when completing the standard processes and there were many good attempts made when questions required explanations to be given, particularly within the pure questions. A common feature of the stronger responses to questions was the inclusion of diagrams.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
Two sequences are defined by $a_1 = 1$ and $b_1 = 2$ and, for $n \ge 1$,
\begin{equation*}
\begin{split}
a_{n+1} & = a_n+ 2b_n \,,
\\
b_{n+1} & = 2a_n + 5b_n \,.
\end{split}
\end{equation*}
Prove by induction that, for all $n \ge 1$,
\[ a_n^2+2a_nb_n - b_n^2 = 1 \,. \tag{$*$}\]
\begin{questionparts}
\item Let $c_n = \dfrac{a_n}{b_n}$. Show that $b_n \ge 2 \times 5^{n-1}$ and use $(*)$ to show that
\[ c_n \to \sqrt 2 -1 \text{ as } n\to\infty\,. \]
\item Show also that $c_n > \sqrt2 -1$ and hence that $\dfrac2 {c_n+1} < \sqrt2 < c_n+1$.
Deduce that $\dfrac{140}{99}< \sqrt{2} < \dfrac{99}{70 }\,$.
\end{questionparts}Claim $a_n^2+2a_nb_n - b_n^2 = 1$ for all $n \geq 1$
Proof: (By induction)
Base case: ($n = 1$).
When $n = 1$ we have $a_1^2 + 2a_1 b_1-b_1^2 = 1^2+2\cdot1\cdot2-2^2 = 1$ as required.
(Inductive step). Now we assume our result is true for some $n =k$, ie $a_k^2+2a_kb_k - b_k^2 = 1$, now consider $n = k+1$
\begin{align*}
&& a_{k+1}^2+2a_{k+1}b_{k+1} - b_{k+1}^2 &= (a_k+2b_k)^2+2(a_k+2b_k)(2a_k+5b_k) - (2a_k+5b_k)^2 \\
&&&= a_k^2+4a_kb_k+4b_k^2 +4a_k^2+18a_kb_k+20b_k^2 - 4a_k^2-20a_kb_k-25b_k^2 \\
&&&= (1+4-4)a_k^2+(4+18-20)a_kb_k +(4+20-25)b_k^2 \\
&&&= a_k^2+2a_kb_k -b_k^2 = 1
\end{align*}
Therefore since our statement is true for $n = 1$ and when it is true for $n=k$ it is true for $n=k+1$ by the POMI it is true for $n \geq 1$
\begin{questionparts}
\item Notice that $b_{n+1} \geq 5 b_n$ and therefore $b_n \geq 5^{n-1} b_1 = 2\cdot 5^{n-1}$, so
\begin{align*}
&& 1 &= a_n^2+2a_nb_n - b_n^2\\
\Rightarrow && \frac1{b_n^2} &= c_n^2 + 2c_n - 1 \\
\to && 0 &= c_n^2 + 2c_n - 1 \quad \text{ as } n\to \infty \\
\end{align*}
This has roots $c = -1 \pm \sqrt{2}$, and since $c_n > 0$ it must tend to the positive value, ie $c_n \to \sqrt{2}-1$
\item Notice that $c_n^2 + 2c_n - 1 > 0$ so either $c_n > \sqrt{2}-1$ or $c_n < -1-\sqrt{2}$, but again, since $c_n > 0$ we must have $c_n > \sqrt{2}-1$.
Therefore $\sqrt{2} < c_n + 1$ and $1+c_n > \sqrt{2} \Rightarrow \frac{1}{1+c_n} < \frac{\sqrt{2}}2 \Rightarrow \frac{2}{1+c_n} < \sqrt{2}$
\begin{array}{c|c|c}
n & a_n & b_n \\ \hline
1 & 1 & 2 \\
2 & 5 & 12 \\
3 & 29 & 70
\end{array}
Therefore $c_3 = \frac{29}{70}$ and so
$\frac{2}{1 + \frac{29}{70}} = \frac{140}{99} < \sqrt{2} < \frac{29}{70} + 1 = \frac{99}{70}$
\end{questionparts}
This question received a relatively high number of attempts, although many did not progress very far and so the average mark for this question was again quite low. Many candidates were very competent with the process of proof by induction, although the fact that the question involved two related sequences caused difficulties for some. There were then a number of good solutions to part (i), but many did not manage to justify the limit of the sequence clearly enough to secure full marks. The difficulty often encountered in the final part was in showing the first result. Often those who successfully achieved this were able to complete the rest of the question successfully.