For sequence'

a n = a n-1 + a n-2

We first assume that a n has a general form of b^n.

Then, we put it into the sequence equation and get a quadratic one.

It will be ok for me if the final solution for the a n be that form.

But normally, if there exists two different roots, the final solution indeed is in the form of (C b^n + D f^n)

where they are arbitrarily assigned.

Why do we assume the incorrect form but getting the correct answer?

Any comments Please!

We do not assume that the solution is b^n at all. We assume that the solution is of the form c_1*b_1^n + c_2*b_2^n: the equation is linear. And then from this knowledge we know that b must satisfy a quadratic equation based upon the information given. All you're doing is taking sloppy shorthand for absolute truth.

Thanks for your help first.

Here I cut a short passage from a lecture from a local university.

I highlighted the part which shows that the prove assumes that the "incorrect" form.

Why can it lead to a more complicated solution?

*may* have solutions of the form, not *does* have. In anycase, it is badly written, but that is not uncommon in textbooks (it for instance states the solution in the degree one case is a^nx_0 and then immediately asserts x_n=a^n is a solution. It is not unless x_0=1

OK. Thanks for telling me that prove is poorly written.

I am going to seek a better one. Do you know where I can find it?