Solution. Since $a_{n+1} a_{n}=4\left(a_{n+1}-1\right), a_{n+1}=\frac{4}{4-a_{n}}$. If we subtract 2 from both sides, we have $a_{n+1}-2=\frac{4}{4-a_{n}}-2=\frac{2\left(a_{n}-2\right)}{4-a_{n}}$. Therefore, $\frac{1}{a_{n+1}-2}=\frac{4-a_{n}}{2\left(a_{n}-2\right)}=\frac{1}{a_{n}-2}-\frac{1}{2}$. Thus, the sequence $\left\{\frac{1}{a_{n}-2}\right\}$ is an arithmetic progression with the first term $\frac{1}{a_{n}-2}=-1$ and common difference $d=-\frac{1}{2}$. Hence, $\frac{1}{a_{n}-2}=-1+(n-1)\left(-\frac{1}{2}\right)=-\frac{n+1}{2}$, from which $a_{n}=\frac{2 n}{n+1}$. Finally,

$$
a_{1} a_{2} \cdot \ldots \cdot a_{n}=\left(2 \cdot \frac{1}{2}\right)\left(2 \cdot \frac{2}{3}\right)\left(2 \cdot \frac{3}{4}\right) \ldots\left(2 \cdot \frac{n}{n+1}\right)=\frac{2^{n}}{n+1}
$$
3A. Find $a_{1} a_{2} \cdot \ldots \cdot a_{n}$, if $a_{1}=1, a_{n+1} a_{n}=4\left(a_{n+1}-1\right)$, for $n \geq 1$.