2007 Paper 3 Q3

Year: 2007
Paper: 3
Question Number: 3

Course: LFM Pure
Section: Proof by induction

Difficulty: 1700.0 Banger: 1469.5

Problem

A sequence of numbers, \(F_1, F_2, \ldots\), is defined by \(F_1=1, F_2=1\), and \[ F_n=F_{n-1}+F_{n-2}\, \quad \text{for \(n\ge 3\)}. \]
  1. Write down the values of \(F_3, F_4, \ldots , F_8\).
  2. Prove that \(F^{\vphantom{2}}_{2k+3}F^{\vphantom{2}}_{2k+1} -F_{2k+2}^2 = -F^{\vphantom{2}}_{2k+2}F^{\vphantom{2}}_{2k}+F_{2k+1}^2\,\).
  3. Prove by induction or otherwise that \(F^{\vphantom{2}}_{2n+1}F^{\vphantom{2}}_{2n-1}-F^2_{2n}=1\,\) and deduce that \(F^2_{2n}+1\,\) is divisible by \(F^{\vphantom{2}}_{2n+1}\,.\)
  4. Prove that \(F^2_{2n-1}+1\,\) is divisible by \(F^{\vphantom{2}}_{2n+1}\,.\)

No solution available for this problem.

Examiner's report
— 2007 STEP 3, Question 3
Above Average

This question was popular. Many solutions to part (ii) were rambling and lacked a sense of direction, even if correct. The induction in (iii) was frequently incorrectly handled and a common error was to replace n by k/2. Part (iv) caused difficulties.

Source: Cambridge STEP 2007 Examiner's Report · 2007-full.pdf
Rating Information

Difficulty Rating: 1700.0

Difficulty Comparisons: 0

Banger Rating: 1469.5

Banger Comparisons: 2

Show LaTeX source
Problem source
A sequence of numbers, $F_1, F_2, \ldots$, is defined by
$F_1=1, F_2=1$, and
\[ 
F_n=F_{n-1}+F_{n-2}\, \quad \text{for $n\ge 3$}.
\]
\begin{questionparts}
\item    Write down the values of $F_3, F_4, \ldots , F_8$.
\item  Prove that $F^{\vphantom{2}}_{2k+3}F^{\vphantom{2}}_{2k+1} -F_{2k+2}^2 = -F^{\vphantom{2}}_{2k+2}F^{\vphantom{2}}_{2k}+F_{2k+1}^2\,$.
\item Prove by induction or otherwise that
$F^{\vphantom{2}}_{2n+1}F^{\vphantom{2}}_{2n-1}-F^2_{2n}=1\,$ and deduce that $F^2_{2n}+1\,$ is divisible by $F^{\vphantom{2}}_{2n+1}\,.$
\item  Prove that  $F^2_{2n-1}+1\,$ is divisible by $F^{\vphantom{2}}_{2n+1}\,.$
\end{questionparts}