For the recurring function $R_n(x)=ax_i+b$, where $x_i$ is the initial number $n$ is the nth iteration $R_n$ is the function value of the nth iteration, $a$ and $b$ are adjustment values.
I've derived the formula that allows us to find the value of the nth iteration of the recurring function, without the long process seen in the example below. However I'm not sure if this equation satisfies all real and complex values of $a, x_i$ and $b$. $$R_n=\frac{((a-1)x_i+b)a^n-b}{a-1}$$
An example is $R(x)=3x_i-7$, where $x_i=3$
$R_1(3)=3(3)-7=2 \rightarrow R_2(2)=3(2)-7=-1 \rightarrow R_3(-1)=3(-1)-7=-10 \rightarrow...\rightarrow R_n(x)=3n-7$
When the values of $a$, $b$, and $x_i$ are substituted into my equation we get the following exponential function. $$R_n=\frac{7-3^n}{2}$$
$R_1=\frac{7-3^1}{2}=2 \rightarrow R_2=\frac{7-3^2}{2}=-1 \rightarrow R_3=\frac{7-3^3}{2}=-10 \rightarrow ... \rightarrow R_n=\frac{7-3^n}{2}$
I have tested many more values, including complex numbers with my equation and had successful results.
My apologies for any confusion, this is the first time I've used this website as well as the first time I've tried to explain my thoughts.
The more common notation for such a linear recurrence would be $\,x_{n+1}=a x_n + b\,$ for $\,n \ge 0\,$, where $\,x_0\,$ is the initial value and $\,a,b$ are constant parameters.
For $\,a=1\,$ the recurrence $\,x_{n+1}=x_n+b\,$ defines the arithmetic progression $\,x_n = x_0 + n b\,$.
Otherwise if $\,a \ne 1\,$ the recurrence relation can be rewritten as: $$ x_{n+1} = ax_n + b \cdot\frac{a-1}{a-1} \quad\iff\quad x_{n+1} + \frac{b}{a-1}= a\left(x_n + \frac{b}{a-1}\right) $$ This means that $\,x_n + \dfrac{b}{a-1}\,$ is a geometric progression with common ratio $\,a\,$, so $x_{n} + \dfrac{b}{a-1} = a^n \left(x_0 + \dfrac{b}{a-1}\right)\,$, and rearranging the terms gives the posted form.
In both cases, the proof works (and the result holds) whether $\,a,b,x_0\,$ are real or complex.