Year: 2004
Paper: 3
Question Number: 6
Course: UFM Pure
Section: Sequences and series, recurrence and convergence
No solution available for this problem.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1503.0
Banger Comparisons: 4
Given a sequence $w_0$, $w_1$, $w_2$, $\ldots\,$, the sequence $F_1$, $F_2$, $\ldots$ is
defined by
$$F_n = w_n^2 + w_{n-1}^2 - 4w_nw_{n-1} \,.$$
Show that
$\;
F_{n}-F_{n-1} = \l w_n-w_{n-2} \r \l w_n+w_{n-2}-4w_{n-1} \r \;
$ for $n \ge 2\,$.
\begin{questionparts}
\item
The sequence $u_0$, $u_1$, $u_2$, $\ldots$
has $u_0 = 1$, and $u_1 = 2$ and satisfies
\[
u_n = 4u_{n-1} -u_{n-2} \quad (n \ge 2)\;.
\]
Prove that
\ $
u_n^2 + u_{n-1}^2 = 4u_nu_{n-1}-3
\; $
for $n \ge 1\,$.
\item
A sequence $v_0$, $v_1$, $v_2$, $\ldots\,$ has $v_0=1$ and satisfies
\begin{equation*}
v_n^2 + v_{n-1}^2 = 4v_nv_{n-1}-3 \quad (n \ge 1). \tag{$\ast$}
\end{equation*}
\makebox[7mm]{(a) \hfill}Find $v_1$ and prove that, for each $n\ge2\,$, either
$v_n= 4v_{n-1} -v_{n-2}$ or $v_n=v_{n-2}\,$.
\makebox[7mm]{(b) \hfill}Show that the sequence, with period 2, defined by
\begin{equation*}
v_n =
\begin{cases}
1 & \mbox{for $n$ even} \\
2 & \mbox{for $n$ odd}
\end{cases}
\end{equation*}
\makebox[7mm]{\hfill}satisfies $(\ast)$.
\makebox[7mm]{(c) \hfill}Find a sequence $v_n$ with period 4
which has $v_0=1\,$, and satisfies~$(\ast)$.
\end{questionparts}