We are asked to evaluate the following limit: $Q = \lim_{x \to 0} \frac{1 - \cos x \cos 2x \cos 3x \dots \cos nx}{x^2}$.

1. Problem Description

We are asked to evaluate the following limit:

$Q = \lim_{x \to 0} \frac{1 - \cos x \cos 2x \cos 3x \dots \cos nx}{x^2}$.

2. Solution Steps

We can rewrite the numerator as:

$1 - \cos x \cos 2x \cos 3x \dots \cos nx = 1 - (1 - (1 - \cos x)) \cos 2x \dots \cos nx$

$= 1 - \cos 2x \cos 3x \dots \cos nx + (1 - \cos x)\cos 2x \cos 3x \dots \cos nx$.

Then, let $P = \cos x \cos 2x \cos 3x \dots \cos nx$. We can write $1 - P$ as

$1 - P = 1 - \cos x + \cos x - \cos x \cos 2x + \cos x \cos 2x - \dots - \cos x \dots \cos nx$

$= (1 - \cos x) + \cos x (1 - \cos 2x) + \cos x \cos 2x (1 - \cos 3x) + \dots + \cos x \dots \cos(n-1)x(1 - \cos nx)$.

Using the small angle approximation for cosine, $\cos x \approx 1 - \frac{x^2}{2}$ as $x \to 0$, we also have $1 - \cos x \approx \frac{x^2}{2}$ as $x \to 0$.

When $x \to 0$, $\cos kx \to 1$ for any integer $k$. Also, $1 - \cos kx \approx \frac{(kx)^2}{2} = \frac{k^2 x^2}{2}$.

Therefore, the numerator becomes:

$1 - P = (1 - \cos x) + \cos x (1 - \cos 2x) + \cos x \cos 2x (1 - \cos 3x) + \dots + \cos x \dots \cos(n-1)x(1 - \cos nx)$

$\approx \frac{x^2}{2} + 1 \cdot \frac{(2x)^2}{2} + 1 \cdot 1 \cdot \frac{(3x)^2}{2} + \dots + 1 \cdot 1 \cdot \dots \cdot 1 \cdot \frac{(nx)^2}{2}$

$= \frac{x^2}{2} + \frac{4x^2}{2} + \frac{9x^2}{2} + \dots + \frac{n^2 x^2}{2}$

$= \frac{x^2}{2} (1 + 4 + 9 + \dots + n^2) = \frac{x^2}{2} \sum_{k=1}^n k^2 = \frac{x^2}{2} \frac{n(n+1)(2n+1)}{6}$.

$Q = \lim_{x \to 0} \frac{\frac{x^2}{2} \frac{n(n+1)(2n+1)}{6}}{x^2} = \lim_{x \to 0} \frac{n(n+1)(2n+1)}{12} = \frac{n(n+1)(2n+1)}{12}$.

3. Final Answer

$\frac{n(n+1)(2n+1)}{12}$