Another on recurrence sequences

[dodo]

How would you go on proving the following conjecture?
Given

$$S_0 = 0, \quad S_1 = 1, \quad S_n = a S_{n-1} + b S_{n-2}$$​

Prove that

$${ S_n }^2 - S_{n-1} S_{n+1} = (-b)^{n-1} \quad (n = 1, 2, 3, ...)$$​

 

[morphism]

Induction seems to work. I put the first couple of steps in white (so highlight if you can't see):S(n+1)^2 - S(n)S(n+2)
= S(n+1)^2 - S(n)(aS(n+1) + bS(n))
= S(n+1)^2 - aS(n)S(n+1) - bS(n)^2

Now think about you can possibly get a (-b)^(n-1) in there.

 

[dodo]

Thanks! Nevermind, I was just doing something like that. Guess I was dense this morning.

Just start from the equation on n-1 and add and substract a . S_n . S_{n-1} to the left-hand side.

Edit: Oh, sorry, first multiply the whole equation by -b, then add and substract the term above.