Let's write the following expression in a simpler form:

$$
\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right) \ldots\left(1-\frac{1}{n^{2}}\right)
$$
I. solution. Each factor is the difference of two squares, and thus can be factored into two further factors. Let's take out the difference of the bases from each factor and consider the product of these:

$$
\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right) \ldots\left(1-\frac{1}{n}\right)=\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \ldots \frac{n-1}{n}=\frac{1}{n}
$$

Similarly, for the product of the sums of the bases:

$$
\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right) \ldots\left(1+\frac{1}{n}\right)=\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \ldots \frac{n+1}{n}=\frac{n+1}{2}
$$

Therefore,

$$
K=\frac{1}{2} \cdot \frac{n+1}{2}=\frac{n+1}{2 n}
$$

For odd $n$, this can still be simplified by 2, but for even $n$, it cannot be simplified, because $n$ and $n-1$ are consecutive natural numbers and thus relatively prime.

Hoffer Anna (Budapest, Hámán K. lg. I. o. t.)

II. solution. By trying out the first few values of $n$, it is easy to notice that the following relationship holds for these:

$$
K=K_{n}=\frac{n+1}{2 n}
$$

We will prove by complete induction that this relationship holds for all further $n$. Indeed, if the relationship is true for $n$, then

$$
\begin{gathered}
K_{n+1}=K_{n}\left(1-\frac{1}{(n+1)^{2}}\right)=\frac{n+1}{2 n} \cdot \frac{(n+1)^{2}-1^{2}}{(n+1)^{2}}= \\
=\frac{(n+1) n(n+2)}{2 n(n+1)^{2}}=\frac{n+2}{2(n+1)}
\end{gathered}
$$

thus, the conjectured regularity is inherited from $n$ to the next natural number. Since the regularity holds for $n=2$, it is valid for all natural numbers greater than 2.

Szidarovszky Klára (Budapest, Ságvári E. lg. I. o. t.)

Note. In fact, we also used complete induction in the first solution, but we could omit the proof, as the possibility and result of simplification were so obvious in the two sub-sequences.