1. ## Sequence Problem

The Sequence $u_n (n = 1,2,3...)$ satisfies the recurrence relation

$u_{n+2} = \frac{u_{n+1}}{u_{n}} (ku_{n} -u_{n+1})$
where k is constant.

If $u_1 = a$ and $u_2 = b$ where a and b are non-zero and $b \neq ka$

Prove by induction that

$u_{2n} = \left( \frac{b}{a} \right ) u_{2n-1}$

$u_{2n+1} = cu_{2n}$

for $n \geq 1$where c is a constant to be found in terms of k, a and b hence find and 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.

i) the sequence $u_n$ is geometric
ii) the sequence $u_n$ has period 2
iii) the sequence $u_n$ has period 4

[OCR STEP(III) 2005, Question 4 ]

I am having some trouble on the induction part of the question, I could do with a tiny bit of guidance, I do not want a full solution just a bit of help if anyone wants to give me a hand.

2. first let's start by checking the base case n = 1:

$\begin{gathered}
\left. {u_{2n} } \right|_{n = 1} = \left. {\left( {\frac{b}
{a}} \right)u_{2n - 1} } \right|_{n = 1} \hfill \\
\Leftrightarrow u_2 = \left( {\frac{b}
{a}} \right)u_1 = b\quad o.k. \hfill \\
\end{gathered}$

now let us assume that the relation holds for some n:

$
1) u_{2n} = \left( {\frac{b}
{a}} \right)u_{2n - 1}
$

now if we prove that the relation holds for n+1, then by the induction hypothesis the relation is true for any n,
thus we have to prove that:

$
u_{2n + 2} = \left( {\frac{b}
{a}} \right)u_{2n + 1}
$

our next step is to use the given recurrence relation, since it is true for every n it must be true in particular for 2n-1:

$
2)\quad u_{2n + 1} = ku_{2n} - \frac{{u_{2n} ^2 }}
{{u_{2n - 1} }}
$

for the same reason it must be true for 2n, so we get:

$
3)\quad u_{2n + 2} = ku_{2n + 1} - \frac{{u_{2n + 1} ^2 }}
{{u_{2n} }}
$

substitute 1) into 2) and then 2) into 3), and you;ll complete your proof...