2005 Paper 3 Q4

Year: 2005
Paper: 3
Question Number: 4

Course: UFM Pure
Section: Sequences and series, recurrence and convergence

Difficulty: 1700.0 Banger: 1457.9

Problem

The sequence \(u_n\) (\(n= 1, 2, \ldots\)) satisfies the recurrence relation \[ u_{n+2}= \frac{u_{n+1}}{u_n}(ku_n-u_{n+1}) \] where \(k\) is a constant. If \(u_1=a\) and \(u_2=b\,\), where \(a\) and \(b\) are non-zero and \(b \ne ka\,\), prove by induction that \[ u_{2n}=\Big(\frac b a \Big) u_{2n-1} \] \[ u_{2n+1}= c u_{2n} \] for \(n \ge 1\), where \(c\) is a constant to be found in terms of \(k\), \(a\) and \(b\). Hence express \(u_{2n}\) and \(u_{2n-1}\) in terms of \(a\), \(b\), \(c\) and \(n\). Find conditions on \(a\), \(b\) and \(k\) in the three cases:
  1. the sequence \(u_n\) is geometric;
  2. \(u_n\) has period 2;
  3. the sequence \(u_n\) has period 4.

No solution available for this problem.

Rating Information

Difficulty Rating: 1700.0

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Banger Rating: 1457.9

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Problem source
The sequence $u_n$ ($n= 1, 2, \ldots$) satisfies the recurrence relation 
\[ 
u_{n+2}= \frac{u_{n+1}}{u_n}(ku_n-u_{n+1})  
\] 
where  $k$ is a constant. 
 
If $u_1=a$ and $u_2=b\,$,  
where $a$ and  $b$ are non-zero and $b \ne ka\,$, prove by induction that 
\[ 
u_{2n}=\Big(\frac b a \Big) u_{2n-1} 
\] 
\[ 
u_{2n+1}= c u_{2n} 
\] 
for $n \ge 1$,  
where $c$ is a constant to be found in terms of $k$, $a$ and $b$. 
Hence express $u_{2n}$ and $u_{2n-1}$ in terms of $a$, $b$, $c$ and $n$. 
 
Find conditions on $a$, $b$ and $k$ in the three cases: 
\begin{questionparts}
\item the sequence $u_n$ is geometric;
\item $u_n$ has period 2; 
\item the sequence $u_n$  has period 4. 
\end{questionparts}