Task B-4.1. Solve the equation

$$
\frac{1}{(2 n-4)!}=\frac{1}{(2 n-3)!}+\frac{8}{(2 n-2)!}
$$
Solution. The number $n$ must satisfy the condition $n \geq 2$.

If we multiply the equation

$$
\frac{1}{(2 n-4)!}=\frac{1}{(2 n-3)!}+\frac{8}{(2 n-2)!}
$$

by $(2 n-2)!$,

we get

$$
(2 n-2)(2 n-3)=2 n-2+8
$$

i.e., $n^{2}-3 n=0$.

The solutions to the last equation are $n=0$ and $n=3$.

However, only the solution $n=3$ satisfies the condition $n \geq 2$.