Scary function and spooky summation!

Relevant wiki: Logarithmic Functions - Problem Solving - Hard

• First note $f(x)=\left((x+1)(x+2)\right)^{\cos (\pi x)}$
• Secondly $\small{\left|\displaystyle\sum_{k=1}^n\log_{10} f(k)\right|=\left|\log_{10} \displaystyle\prod_{k=1}^n f(k)\right|}=\mathfrak T$
• Also, $\cos k\pi=\begin{cases} -1&\text{if k is odd }\\ 1&\text{if k is even}\end{cases}$

Hence, $\mathfrak T=\left|\log_{10}\left(\dfrac{1}{2\cdot \cancel{3}}\times \cancel{3}\cdot \cancel{4}\times \dfrac 1{\cancel{4}\cdot\cancel{ 5}}\times \cdots\right)\right|$

The last term will be $((n+1)(n+2))^{\cos n\pi}$ in which $n+1$ will get cancelled (by noticing the pattern) while the term $n+2$ will appear in denominator or numerator will depend on $n$ being odd or even.Hence,

$\mathfrak T=\begin{cases}\left|\log_{10}\left(\dfrac{n+2}{2}\right)\right|&\text{if n is even}\\\left|\log_{10}\left(\dfrac{1}{2(n+2)}\right)\right|&\text{if n is odd}\end{cases}$

Hence, equation turns to :

$10^{\pm 1}= \begin{cases}\dfrac{n+2}{2}&\text{if n is even}\\\dfrac{1}{2(n+2)}&\text{if n is odd}\end{cases}$

Since $n$ is an integer, in first case we get $\dfrac{n+2}{2}=10$ ( $n$ is even) while in second case we get $\dfrac 1{2(n+2)}=\dfrac 1{10}$ ( $n$ is odd). So that we get

$n=18,3$ in the respective cases. Their product is $\large 18\times 3=\boxed{\color{#007fff}{54}}$ .