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

$\displaystyle u_{n+2} = \frac{u_{n+1}}{u_{n}} (ku_{n} -u_{n+1})$

where k is constant.

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

Prove by induction that

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

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

for $\displaystyle n \geq 1$where c is a constant to be found in terms of k, a and b hence find and express $\displaystyle u_{2n}$ and $\displaystyle u_{2n-1}$ in terms of a,b,c and n

find conditions on a, b and k in the three cases.

i) the sequence $\displaystyle u_n$ is geometric

ii) the sequence $\displaystyle u_n$ has period 2

iii) the sequence $\displaystyle u_n$ 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.