## Task B-3.2.

Determine all even natural numbers $n$ for which

$$
\left|\log _{0.5} 20^{\cos \pi}+\log _{0.5} 30^{\cos 2 \pi}+\log _{0.5} 42^{\cos 3 \pi}+\cdots+\log _{0.5}\left(n^{2}+7 n+12\right)^{\cos n \pi}\right|=1
$$
## Solution.

Simplify the given expression.

$$
\begin{array}{cc}
\left|\log _{0.5}\left(20^{\cos \pi} \cdot 30^{\cos 2 \pi} \cdot 42^{\cos 3 \pi} \cdots\left(n^{2}+7 n+12\right)^{\cos n \pi}\right)\right|=1 & \\
\left|\log _{0.5}\left(20^{-1} \cdot 30^{1} \cdot 42^{-1} \cdots\left(n^{2}+7 n+12\right)^{1}\right)\right|=1 & 1 \text { point } \\
\left|\log _{0.5}\left((4 \cdot 5)^{-1} \cdot(5 \cdot 6) \cdot(6 \cdot 7)^{-1} \cdots \cdots(n+3)(n+4)\right)\right|=1 & 1 \text { point } \\
\left|\log _{0.5}\left(\frac{1}{4 \cdot 5} \cdot(5 \cdot 6) \cdot \frac{1}{6 \cdot 7} \cdots(n+3)(n+4)\right)\right|=1 & 1 \text { point } \\
\left|\log _{0.5}\left(\frac{1}{4} \cdot(n+4)\right)\right|=1 \Longrightarrow \log _{0.5}\left(\frac{1}{4} \cdot(n+4)\right)= \pm 1 & 1 \text { point }
\end{array}
$$

From $\frac{n+4}{4}=\frac{1}{2}$, we get $n=-2$, but this is not a natural number.

From $\frac{n+4}{4}=2$, we get the solution $n=4$.