Vtyjkyuk

I started with the recurrence equation $spacespace x_n+1=ax_n+b space space$ and turned it into the formula at the bottom to allow us to find the value of any nth term in the form $space space x_n=f(n)$

Here is how I derived my formula for $f(n)$:

$$x_0=x$$
$$x_1=ax+b$$
$$x_2=a^2x+ab+b$$
$$x_3=a^3x+a^2b+ab+b$$
$$x_n=a^nx+a^n-1b+a^n-2b...a^2b+ab+b$$
Refer to the geometric progression equation:
$$sum_k=1^na^k-1b=fracb(1-a^n)1-a$$
$$x_n=a^nx+fraca^nb-ba-1$$
$$x_n=fraca^n+1x-a^nx+a^nb-ba-1$$
$$x_n=fraca^n(ax-x+b)-ba-1$$
Final simplification and substitution of $x_0=x$
$$x_n=fraca^n((a-1)x_0+b)-ba-1$$
Side Note: This idea originally came from problems I encountered in the Collatz Conjecture, eventually leading to create a general formula. Could this potentially be helpful to have other than my own personal use?

You can make it easier. Considering $$ x_n+1=a,x_n+b $$ let $x_n=y_n+c$ to make
$$y_n+1+c=a, y_n+ac+b$$ that is to say
$$y_n+1=a, y_n+(ac+b-c)$$ Choose $c$ such that $ac+b-c=0$ and you are back to the classical geometric sequence. Solve it for $y_n$ and go back to $x_n$.

This is a linear recurrence with constant coefficients and it can also be solved similarly to a linear ODE.

The homogeneous equation is

$$x_n+1=ax_n$$ and obviously has the general solution

$$x_n=ca^n.$$

Now a particular solution of the non-homegeneous equation

$$x_n+1=ax_n+b$$ is given by a constant, let $d$, such that

$$d=ad+b.$$

We have

$$x_n=ca^n+frac b1-a$$ and we plug the initial condition $x=x_0$,

$$x_0=c+frac b1-a$$ and finally

$$x_n=a^nx_0+frac1-a^n1-ab.$$

Other method:

The relation $x_n+1=ax_n$ hints the transformation $x_n=a^ny_n$, leading to the modified recurrence

$$y_n+1=y_n+ba^-n,$$ which is solved by the summation of a geometric series

$$y_n=y_0+bsum_k=1^na^-k=y_0+a^-1frac1-a^-n1-a^-1.$$

Then

$$x_n=a^nx_0+fraca^n-1a-1b.$$