If a sequence is such that the next term is given by $a_n*a +b$, what is the general formula of the given sequence?
e.g.: $-$ for $a=4,\;b=6$, the sequence is:
$1,\quad1*4+6=10,\quad10*4+6=46,\quad46*4 +6=190,\;\dots$
What would the formula for the general term be?
(I'm not necessarily asking specifically for this example but rather for a general sequence with the next term defined by $a_n*a +b$).
Empirically,
$$t_1=at_0+b,\\t_2=a^2t_0+ab+b,\\t_3=a^3t_0+a^2b+ab+b,\\t_4=a^4t_0+a^3b+a^2b+ab+b,\\\cdots$$
You easily see the pattern, $a^nt_0$ plus $b$ times the sum of the geometric series $a^k$.
This yields the formula
$$t_n=a^nt_0+b\frac{a^n-1}{a-1}.$$
As we can check,
$$t_0=a^0t_0+b\frac{a^0-1}{a-1}$$ is true, and
$$at_n+b=aa^nt_0+ab\frac{a^n-1}{a-1}+b=a^{n+1}t_0+b\frac{a^{n+1}-a+a-1}{a-1}=t_{n+1}.$$
If the first element of the sequence is $s_0$, when you compute the first terms of your sequence it won't be difficult to guess that the $n$-th term of that is $$s(n)=s_0a^n+b\left(\sum_{i=0}^{n-1}a^i\right)$$ then go by induction.
If the initial term is $u_0$ and if $a\ne 1$, calculating some terms, you can conjecture the formula should be $$u_n=a^n u_0+\frac{a^n-1}{a-1}\,b\qquad(n\ge 1),$$ which is easy to prove by induction.
You want to find a formla for a sequence define by $u_{n+1}=au_n+b$ for all $n\geq 0$.
The idea is to try to reduce to geometric sequences:
We want to find $\alpha$ such that the sequence $(u_n-\alpha)_n$ is geometric of parameter $a$ (it is the most natural choice, because if $b=0$, we already are in this case).
So we want to have $u_{n+1}-\alpha=a(u_n-\alpha)=au_n+b-\alpha$. So, $\alpha=\dfrac{b}{1-a}$ does the job, provided $a\neq 1$.
Assume that $a\neq 1$, otherwise we have $u_{n+1}=u_n+b$ and this is a case we know how to solve.
Then set $v_n=u_n-\dfrac{b}{1-a}$. Then $v_{n+1}=av_n$ for all $n\geq 0$, so $v_n=v_0a^n=(u_0-\dfrac{b}{1-a})a^n$.
Therefore $u_n=(u_0-\dfrac{b}{1-a})a^n+\dfrac{b}{1-a}$ for all $n\geq 0$.
$$t_{n+1}=at_n+b$$ is an affine recurrence, and we can decompose in "homogeneous" and "particular" solutions, as follows.
If we hypothetize that $t_n$ is a constant, let $t$, we have
$$t=at+b$$ or $$t=\frac b{1-a}.$$
Now by subtraction of the first two equations,
$$t_{n+1}-t=a(t_n-t)$$ and this recurrence defines a geometric series. So
$$t_n-t=a^n(t_0-t)$$ and
$$t_n=a^n\left(t_0-\frac b{1-a}\right)+\frac b{1-a}.$$
You have recursion $$x_{n+1} = a x_n + b \qquad a \neq 1, \quad b\neq 0$$ Note that $$ x_{n+1} - \frac{b}{1-a} = a x_n + b - \frac{b}{1-a} = a x_n - \frac{ab}{1-a} = a\big(x_{n+1} - \frac{b}{1-a}\big)$$ which means that $y_n = x_n - \frac{b}{1-a} $ is a geometric series: $$ y_{n+1} = a y_n$$ It can be solved $$ y_n = a^n y_0$$ and you get $$ x_n = y_n + \frac{b}{1-a} = a^n(x_0 - \frac{b}{1-a}) + \frac{b}{1-a}$$ $$ x_n = a^n x_0 + b\frac{1-a^n}{1-a}$$
