Easy method for sequence equation simplifying I have a problem with one of my questions.

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*Check the convergence of the sequence defined by $$u_{n+2}=
\frac{1}{2}(u_n +u_{n+1}), \;\;\;u_1 =a, \;u_2 =b.$$
So in there my tutor got the expression
$$u_{n+2}=u_{n+1} + \frac{(-1)^n}{2^n}(b-a)$$
suddenly, and proved it by induction method.
After a considerable time I realized, we can write this format by getting $(u_{n+2}-u_{n+1})$ values for $u_3, u_4, \dots$ so on.
I want to learn is there any easy method to understand for get such an equation. Thanks in advance :)
 A: You find its characteristic equation (I'm unsure if it's called this way in other parts of the world): $2\lambda^2 - \lambda - 1 = 0 \Rightarrow \lambda = 1 \wedge \lambda = \frac{-1}{2}$
Then $u_n = A \cdot (1)^n + B \cdot (\frac{-1}{2})^n = A + B \cdot (\frac{-1}{2})^n, \forall n \in \mathbb{N}^{+}$
$u_1 = A - \frac{B}{2} = a$
$u_2 = A + \frac{B}{4} = b$
$\Rightarrow B = \frac{4(b - a)}{3}$
$u_{n + 1} - u_{n} = A + B \cdot (\frac{-1}{2})^{n+1}- A - B \cdot (\frac{-1}{2})^n = \frac{4(b-a)}{3} \cdot (\frac{-1}{2})^n \cdot (\frac{-1}{2} - 1) = \frac{4(b-a)}{3} \cdot (\frac{-1}{2})^n \cdot (\frac{-3}{2}) = 4(b-a) \cdot (\frac{-1}{2})^{n+1} = (b - a) \cdot (\frac{-1}{2})^{n - 1}$
Which also means: $u_{n + 2} - u_{n + 1} = (b - a) \cdot (\frac{-1}{2})^{n}$, and this is what we wanted.
If you're unfamiliar with what written above, I'd recommend reading a book about this topic, I think it may be called "Finding the closed form of linear recurrence sequences". Unfortunately, I've got no book in English about it. If you are Vietnamese, by any chance, please inform me, I'll recommend you a few books that I have read.
