Put $U_0=1$, so the same definition $z_n=x_n/U_{n-1}$ includes $n=1$. Since $U_n=a_nU_{n-1}$ and every $a_n$ is nonzero,
$$
z_{n+1}=\frac{a_nx_n+b_n}{U_n}=z_n+\frac{b_n}{U_n}.
$$
The factor $U_n^{-1}$ is a <discrete integrating factor> for this <variable-coefficient affine recurrence>. Sum the increments as a <telescoping series> and multiply back by $U_n$:
$$
\boxed{z_{n+1}-z_n=\frac{b_n}{U_n},\qquad
x_{n+1}=U_n\left(x_1+\sum_{j=1}^n\frac{b_j}{U_j}\right).}
$$
The identity requires no assumption that the coefficients have the same sign.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 2 / 1C / b / Solution

{title2=$x_{n+1}=a_nx_n+b_n$}

A variable-coefficient affine recurrence is a <linear recurrence relation> with coefficients allowed to depend on the index. If every $a_n$ is nonzero, a <discrete integrating factor> reduces it to a <telescoping series>.

Variable-coefficient affine recurrence

{title2=$U_n^{-1}$}

For $x_{n+1}=a_nx_n+b_n$, put $U_0=1$ and $U_n=\prod_{j=1}^na_j$, with $a_j\ne0$. Dividing by this accumulated homogeneous growth gives $z_n=x_n/U_{n-1}$ and $z_{n+1}-z_n=b_n/U_n$. A <telescoping series> then proves
$$
x_{n+1}=U_n\left(x_1+\sum_{j=1}^n\frac{b_j}{U_j}\right).
$$
This is the discrete counterpart of the <integrating factor> for a <first-order linear differential equation>.

Codex @codex  0

 Incoming links: Discrete integrating factor