I'm reading an online note by a teacher in my country on finding the general formula of different types of sequences. The note starts by discussing basic sequences like the arithmetic sequence and the geometric sequence. The associated formulas and their derivations are well-known to high school students in my country, which is the target audience of this note. It then discusses some advanced forms that are extensions of the arithmetic and geometric sequences and shows how to derive their general formula. It seems to use the same technique: write $u_n = v_n + f(n)$; As such, $u_{n+1} = v_{n+1} + f(n+1) $; based on the given relationship between $u_n$ and $u_{n+1}$, we could derive a relationship between $v_{n+1}$ and and $v_n$ . The key is the choice of $f(n)$ to make $v_n$ a known sequence.
For example, on discussing this sequence $$\begin{cases} u_1 = a, \\ u_{n+1} = qu_n + d, & \text{q, d ≠ 0 & q ≠ 1} \end{cases}$$ the author simply said, write $u_n = v_n + \frac{d}{1-q}$
$⟹ v_{n+1} + \frac{d}{1-q} = q(v_n + \frac{d}{1-q}) + d$
$⟺ v_{n+1} = qv_n$
As such, $v_n$ is a geometric sequence with factor $q$, so $v_n = v_1q^{n-1} = (a - \frac{d}{1-q}).q^{n-1}$
So $u_n = (a - \frac{d}{1-q}).q^{n-1} + \frac{d}{1-q}$
I struggle to see how I could have come up with this idea myself. How to know that I should write $u_n = v_n + f(n)$? How to know that I should choose $f(n) = \frac{d}{1-q}$ in this case?
Please advise! Thank you very much!
Hint
$$ u_{n+1} = q\,u_n + d$$
Let $u_n=v_n+A$ and replace $$v_{n+1}=q\,v_n+ A(q-1)+d$$
Do you see it better ?